Arithmetic Progression ICSE Class 10

This page provides detailed ICSE Class 10 Maths Arithmetic Progression Solutions from the Selina Concise Mathematics textbook. An Arithmetic Progression, or A.P., is a special sequence of numbers where each term after the first is found by adding a fixed number, called the common difference, to the preceding term. In this chapter, you will master the core concepts of A.P., including how to find the general term (the nth term), calculate the sum of a certain number of terms, and insert arithmetic means between two given numbers. Understanding these formulas is crucial for solving a variety of problems, from simple sequences to more complex real-world applications. Our step-by-step solutions will guide you through each concept clearly.

If you are looking for clear, step-by-step answers for any question in the Arithmetic Progression chapter, you have come to the right place. We have solved all 115 questions from Exercise 10(A), 10(B), 10(C), 10(D), Test Yourself, and the Case-Study Based Questions. Each solution is worked out using the same method and format that the ICSE board expects in your exams, ensuring you learn the correct approach. Here you will find reliable, easy-to-follow solutions to help you verify your answers and master every concept in this chapter.

Exercise 10(A)

Question 1(a)

The first term and the common difference of an A.P. are 8 and -5 respectively. The A.P. is :

(a) 8, 13, 18, 23, 28, ………
(b) 8, 3, -2, -7, …….
(c) -5, 3, 11, 19, 27, ……..
(d) -5, -13, -21, -29, …….

Answer: (b) 8, 3, -2, -7, …….

We have the first term $a = 8$ and the common difference $d = -5$ . The terms of the arithmetic progression (A.P.) are calculated as $a$ , $a + d$ , $a + 2d$ , $a + 3d$ , and so on.

∴ The sequence becomes:

8, 8 + (-5), 8 + 2(-5), 8 + 3(-5), \ldots

Simplifying, we get:

8, 8 - 5, 8 - 10, 8 - 15, \ldots

This results in the sequence:

8, 3, -2, -7, \ldots

Hence, Option 2 is the correct option.

Question 1(b)

Is -8, -8, -8, -8, ….. an A.P.?

(a) no
(b) yes
(c) may be
(d) none of the above

Answer: (b) yes

The sequence given is indeed an arithmetic progression. Here, the common difference is calculated as follows: (-8 – (-8) = -8 + 8 = 0). Since the common difference is 0, the sequence qualifies as an A.P.

Hence, Option 2 is the correct option.

Question 1(c)

The 15^th term of the A.P. 3, 0, -3, -6, …… is :

(a) -42
(b) 39
(c) 42
(d) -39

Answer: (d) -39

In the arithmetic progression given by 3, 0, -3, -6, ……, we identify the first term as $a = 3$ . The common difference $d$ is calculated as $0 - 3 = -3$ .

To find the 15th term $a_{15}$ , we use the formula for the $n$ -th term of an arithmetic progression:

a_n = a + (n - 1) \times d

Substituting the known values, we have:

a_{15} = 3 + (15 - 1) \times (-3)

This simplifies to:

a_{15} = 3 + 14 \times (-3)

a_{15} = 3 - 42

a_{15} = -39

Hence, Option 4 is the correct option.

Question 1(d)

The 24th term of an A.P. exceeds its 19^th term by 10, its common difference is :

(a) 5
(b) 2
(c) 10
(d) 1

Answer: (b) 2

Consider the first term of the arithmetic progression (A.P.) as $a$ and the common difference as $d$ .

Using the formula for the $n^{th}$ term of an A.P., we have:

a_n = a + (n - 1)d

For the 24th term:

a_{24} = a + (24 - 1)d = a + 23d

For the 19th term:

a_{19} = a + (19 - 1)d = a + 18d

According to the given condition, the 24th term is 10 more than the 19th term:

a_{24} - a_{19} = 10

Substituting the expressions for $a_{24}$ and $a_{19}$ :

(a + 23d) - (a + 18d) = 10

Simplifying, we get:

a - a + 23d - 18d = 10

5d = 10

Solving for $d$ :

d = \frac{10}{5} = 2

Hence, Option 2 is the correct option.

Question 1(e)

In the A.P. 8, 13, 18, ……., the n^th term is 83, then n is equal to :

(a) 14
(b) 16
(c) 13
(d) 15

Answer: (b) 16

In this arithmetic progression, we have:

First term ((a) = 8)

Common difference ((d)) is calculated as follows:

$d = 13 - 8 = 5$ .

Using the formula for the nth term of an arithmetic progression:

a_n = a + (n - 1) \, d

Substitute the given values:

83 = 8 + 5(n - 1)

Simplifying the equation:

83 = 8 + 5n - 5

83 = 5n + 3

Subtract 3 from both sides:

83 - 3 = 5n

80 = 5n

Divide both sides by 5 to solve for $n$ :

$n = \dfrac{80}{5} = 16$ .

Hence, option 2 is the correct option.

Question 2

The n^th term of a sequence is (2n – 3), find the fifteenth term.

Answer:

We have the formula for the n-th term of the sequence as $a_n = 2n - 3$ .

To find the fifteenth term, substitute $n = 15$ into the formula:

a_{15} = 2(15) - 3

Calculate step-by-step:

a_{15} = 30 - 3 = 27

Hence, fifteenth term = 27.

Question 3

If the p^th term of an A.P. is (2p + 3); find the A.P.

Answer:

We know that the p^th term of the arithmetic progression (A.P.) is given by:

⇒ a_p = 2p + 3

To find the first few terms of the A.P., we substitute specific values for p.

For the first term (p = 1):

⇒ a_1 = 2(1) + 3 = 5

For the second term (p = 2):

⇒ a_2 = 2(2) + 3 = 7

For the third term (p = 3):

⇒ a_3 = 2(3) + 3 = 9

Therefore, the arithmetic progression is 5, 7, 9 …….

Question 4

Find the 24^th term of the sequence :

12, 10, 8, 6 ……..

Answer:

First, observe the sequence: 12, 10, 8, 6, … Each term is decreasing by 2. Specifically, 10 – 12 = -2, 8 – 10 = -2, and 6 – 8 = -2.

Thus, this is an arithmetic progression (A.P.) with a common difference of -2.

The formula for the n^th term of an A.P. is:

a_n = a + (n - 1) \, d

where $a$ is the first term. Here, $a = 12$ and $d = -2$ .

To find the 24^th term, use:

a_{24} = 12 + (24 - 1)(-2)

This simplifies to:

a_{24} = 12 + 23(-2) = 12 - 46 = -34

Therefore, the 24^th term of the sequence is -34.

Question 5

Find the 30^th term of the sequence :

\dfrac{1}{2}, 1, \dfrac{3}{2}, .........

Answer:

Observe that $1 - \dfrac{1}{2} = \dfrac{1}{2}$ and $\dfrac{3}{2} - 1 = \dfrac{1}{2}$ .

∴ This sequence forms an arithmetic progression (A.P.) with the common difference $d = \dfrac{1}{2}$ .

The formula for the n^th term of an A.P. is:

⇒ a_n = a + (n – 1)d, where $a$ is the first term.

Thus, to find the 30^th term:

a_{30} = \dfrac{1}{2} + (30 - 1) \times \dfrac{1}{2}

= \dfrac{1}{2} + \dfrac{29}{2}

= \dfrac{30}{2}

= 15.

Therefore, the 30^th term of the sequence is 15.

Question 6

Find the 100^th term of the sequence :

\sqrt{5}, 2\sqrt{5}, 3\sqrt{5}, .......

Answer:

Observe the sequence: $\sqrt{5}, 2\sqrt{5}, 3\sqrt{5}, \ldots$ .

Notice that the difference between the second term and the first term is $2\sqrt{5} - \sqrt{5} = \sqrt{5}$ , and the difference between the third term and the second term is $3\sqrt{5} - 2\sqrt{5} = \sqrt{5}$ . This consistent difference confirms that the sequence is an arithmetic progression (A.P.).

For this A.P., the first term $a$ is $\sqrt{5}$ and the common difference $d$ is $\sqrt{5}$ .

To find the 100th term $a_{100}$ , use the formula for the $n$ th term of an A.P.:
$a_n = a + (n - 1)d$

Substituting the known values, we have:
$a_{100} = \sqrt{5} + (100 - 1) \times \sqrt{5}$

This simplifies to:
$a_{100} = \sqrt{5} + 99\sqrt{5}$

Combining like terms, we get:
$a_{100} = 100\sqrt{5}.$

Thus, the 100th term of the sequence is $100\sqrt{5}$ .

Question 7

Find the 50^th term of the sequence :

\dfrac{1}{n}, \dfrac{n + 1}{n}, \dfrac{2n + 1}{n}, ......

Answer:

To determine the 50th term of the sequence, we first need to confirm if the sequence is an arithmetic progression (A.P.).

Calculate the difference between the second and first terms:

\dfrac{n + 1}{n} - \dfrac{1}{n} = \dfrac{n}{n} = 1

Similarly, calculate the difference between the third and second terms:

\dfrac{2n + 1}{n} - \dfrac{n + 1}{n} = \dfrac{n}{n} = 1

Notice that both differences are equal to 1. ∴ the sequence is indeed an A.P. with a common difference (d) of 1.

The formula for the n-th term of an A.P. is given by:

a_n = a + (n - 1)d

Here, $a$ is the first term of the sequence, which is $\dfrac{1}{n}$ .

To find the 50th term:

a_{50} = \dfrac{1}{n} + (50 - 1) \times 1

= \dfrac{1}{n} + 49

Hence, the 50th term of the sequence is $\dfrac{1}{n} + 49$ .

Question 8

Is 402 a term of the sequence :

8, 13, 18, 23, ……..?

Answer:

Observe the differences between consecutive terms: 13 – 8 = 5, 18 – 13 = 5, and 23 – 18 = 5. This confirms that the sequence is an arithmetic progression (A.P.) with a common difference of 5.

The formula for the n-th term of an A.P. is:

⇒ a_n = a + (n – 1)d, where ‘a’ is the first term.

Suppose 402 is the n-th term of this sequence:

⇒ 402 = a + (n – 1)d

Substituting the values, we have:

⇒ 402 = 8 + (n – 1)5

Simplifying further:

⇒ 402 = 8 + 5n – 5

⇒ 402 = 5n + 3

⇒ 399 = 5n

⇒ n = $\dfrac{399}{5}$

Since ‘n’ must be a whole number, and here it is a fraction,

Hence, 402 is not a term of the sequence.

Question 9

Find the common difference and 99^th term of the arithmetic progression :

7\dfrac{3}{4}, 9\dfrac{1}{2}, 11\dfrac{1}{4}, .......

Answer:

Consider the arithmetic sequence given: $\dfrac{31}{4}, \dfrac{19}{2}, \dfrac{45}{4}, \ldots$ .

To find the common difference, subtract the first term from the second term:

Common difference $= \dfrac{19}{2} - \dfrac{31}{4}$

Convert $\dfrac{19}{2}$ to a fraction with the same denominator as $\dfrac{31}{4}$ :

= \dfrac{38}{4} - \dfrac{31}{4}

= \dfrac{38 - 31}{4}

= \dfrac{7}{4}

= 1\dfrac{3}{4}

The formula for the $n^{th}$ term of an arithmetic progression is given by:

$a_n = a + (n - 1)d$ , where $a$ is the first term and $d$ is the common difference.

To find the $99^{th}$ term:

a_{99} = \dfrac{31}{4} + (99 - 1) \times \dfrac{7}{4}

= \dfrac{31}{4} + \dfrac{98 \times 7}{4}

= \dfrac{31}{4} + \dfrac{686}{4}

= \dfrac{717}{4}

$= 179\dfrac{1}{4}$ .

Therefore, the $99^{th}$ term of the sequence is $179\dfrac{1}{4}$ and the common difference is $1\dfrac{3}{4}$ .

Question 10(i)

How many terms are there in the series :

4, 7, 10, 13, ……….., 148 ?

Answer:

Observe that the differences between consecutive terms are consistent: 7 – 4 = 3, 10 – 7 = 3, and 13 – 10 = 3. This shows that the sequence forms an Arithmetic Progression (A.P.) with a common difference of 3, and the last term is 148.

The formula for the n-th term of an A.P. is:

a_n = a + (n - 1)d

Substituting the known values:

148 = 4 + (n - 1) \times 3

Simplifying further:

148 = 4 + 3n - 3

148 = 3n + 1

Subtract 1 from both sides:

148 - 1 = 3n

3n = 147

Divide both sides by 3:

n = 49.

Hence, no. terms in the series = 49.

Question 10(ii)

How many terms are there in the series :

0.5, 0.53, 0.56, …….., 1.1 ?

Answer:

Observe that the difference between consecutive terms, 0.53 – 0.5 and 0.56 – 0.53, is 0.03. Thus, the sequence is an arithmetic progression (A.P.) with a common difference of 0.03, and the final term given is 1.1.

The formula for the $n^{th}$ term of an A.P. is:

a_n = a + (n - 1)d

Substituting the known values:

1.1 = 0.5 + (n - 1) \times 0.03

Simplifying, we have:

1.1 = 0.5 + 0.03n - 0.03

1.1 = 0.47 + 0.03n

Subtracting 0.47 from both sides:

1.1 - 0.47 = 0.03n

0.63 = 0.03n

Dividing both sides by 0.03:

n = 21

Therefore, the series contains 21 terms.

Question 10(iii)

How many terms are there in the series :

$\dfrac{3}{4}, 1, 1\dfrac{1}{4}, ........., 3$ ?

Answer:

Observe that $1 - \dfrac{3}{4} = \dfrac{1}{4}$ and $1\dfrac{1}{4} - 1 = \dfrac{1}{4}$ .

This indicates our series forms an arithmetic progression (A.P.) with a common difference of $\dfrac{1}{4}$ . The last term of this series is 3.

The formula for the $n^{th}$ term of an A.P. is:

a_n = a + (n - 1) d

Applying this to our series:

\Rightarrow 3 = \dfrac{3}{4} + (n - 1) \times \dfrac{1}{4}

Simplifying the equation:

\Rightarrow \dfrac{3 + n - 1}{4} = 3

\Rightarrow n + 2 = 12

\Rightarrow n = 10.

Therefore, the number of terms in the series is 10.

Question 11

Which term of the A.P. 1, 4, 7, 10, ……. is 52 ?

Answer:

Consider the given arithmetic progression.

The common difference can be calculated as follows: $4 - 1 = 3$ .

The first term of the sequence is 1.

Assume that 52 is the $n^{th}$ term of the sequence.

∴ (52 = 1 + (n – 1) \times 3)

⇒ $52 = 1 + 3n - 3$

⇒ $52 = 3n - 2$

⇒ $54 = 3n$

⇒ $n = 18$ .

Hence, 52 is 18th term of the A.P.

Question 12

If 5^th and 6^th terms of an A.P. are respectively 6 and 5, find the 11^th term of the A.P.

Answer:

In an arithmetic progression (A.P.), the formula for the $n^{th}$ term is given by:

a_n = a + (n - 1)d

We know:

For the $5^{th}$ term, $a_5 = 6$ :

a + 4d = 6 \quad \text{...(i)}

And for the $6^{th}$ term, $a_6 = 5$ :

a + 5d = 5 \quad \text{...(ii)}

Now, let’s subtract equation (i) from equation (ii):

(a + 5d) - (a + 4d) = 5 - 6

This simplifies to:

a - a + 5d - 4d = -1

Thus, we find:

d = -1

Next, substitute $d = -1$ back into equation (i):

6 = a + 4(-1)

This simplifies to:

6 = a - 4

Solving for $a$ , we get:

a = 10

Now, let’s determine the $11^{th}$ term of the A.P.:

a_{11} = 10 + (11 - 1)(-1)

Simplifying further:

= 10 + 10(-1)

= 10 - 10 = 0

Hence, the $11^{th}$ term of the sequence is 0.

Question 13

If t~n represents n^th term of an A.P., t~2 + t~5 – t~3 = 10 and t~2 + t~9 = 17, find its first term and its common difference.

Answer:

The formula for the $n^{th}$ term of an arithmetic progression (A.P.) is given by:

t_n = a + (n - 1)d

Let’s use this formula to solve the equations given.

First, consider the equation $t_2 + t_5 - t_3 = 10$ :

[a + (2 - 1)d] + [a + (5 - 1)d] - [a + (3 - 1)d] = 10

This simplifies to:

(a + d) + (a + 4d) - (a + 2d) = 10

Combining like terms, we get:

a + 3d = 10 \quad \text{...(i)}

Now, take the second equation $t_2 + t_9 = 17$ :

[a + (2 - 1)d] + [a + (9 - 1)d] = 17

This simplifies to:

a + d + a + 8d = 17

Combine the terms:

2a + 9d = 17 \quad \text{...(ii)}

To eliminate $a$ , multiply equation (i) by 2 and subtract from equation (ii):

2a + 9d - 2(a + 3d) = 17 - 2(10)

Simplifying gives:

2a + 9d - 2a - 6d = 17 - 20

3d = -3

⇒ $d = -1$ .

Now, substitute $d = -1$ back into equation (i):

a + 3(-1) = 10

a - 3 = 10

a = 10 + 3 = 13

Thus, the first term is 13 and the common difference is -1.

Question 14

Find the 10^th term from the end of the A.P. 4, 9, 14, ……., 254.

Answer:

To determine the 10 $^{th}$ term from the end of the arithmetic progression (A.P.) given by 4, 9, 14, …, 254, we use the concept that the $r^{th}$ term from the end is equivalent to the ((n – r + 1)^{th}) term from the start, where $n$ is the total number of terms.

First, identify the common difference $d$ of the A.P., which is $9 - 4 = 5$ , and the first term $a = 4$ .

Next, apply the formula for the $n^{th}$ term of an A.P., (a_n = a + (n – 1)d), to find $n$ :

254 = 4 + (n - 1) \times 5

Simplify the equation:

254 = 4 + 5n - 5

254 = 5n - 1

Add 1 to both sides:

255 = 5n

Divide by 5:

n = 51

Thus, the sequence contains 51 terms. The 10 $^{th}$ term from the end is the ((51 – 10 + 1)^{th}), or 42 $^{nd}$ term from the beginning.

Now, substitute into the $n^{th}$ term formula again:

a_{42} = 4 + (42 - 1) \times 5

Calculate:

a_{42} = 4 + 41 \times 5

a_{42} = 4 + 205

a_{42} = 209

Hence, the 10 $^{th}$ term from the end is 209.

Question 15

Determine the arithmetic progression whose 3^rd term is 5 and 7^th term is 9.

Answer:

The formula for the n-th term of an arithmetic progression (A.P.) is:

a_n = a + (n - 1)d

Here, the 3rd term is provided as 5, so we have:

a_3 = a + (3 - 1)d

This simplifies to:

5 = a + 2d

Thus, we have the equation:

a + 2d = 5 \quad \text{...(i)}

Similarly, the 7th term is given as 9:

a_7 = a + (7 - 1)d

Which simplifies to:

9 = a + 6d

Thus, we have another equation:

a + 6d = 9 \quad \text{...(ii)}

To find the common difference $d$ , subtract equation (i) from equation (ii):

(a + 6d) - (a + 2d) = 9 - 5

This results in:

4d = 4

Solving for $d$ , we get:

d = 1

Now, substitute $d = 1$ back into equation (i):

a + 2(1) = 5

Simplifying gives:

a + 2 = 5

So, $a = 3$ .

Thus, the arithmetic progression is:

a, (a + d), (a + 2d), (a + 3d), \ldots

Which translates to:

3, 4, 5, 6, 7, \ldots

Hence, A.P. = 3, 4, 5, 6, 7, \ldots

Question 16

Find the 31^st term of an A.P. whose 10^th term is 38 and 16^th term is 74.

Answer:

The formula for the $n^{th}$ term of an arithmetic progression (A.P.) is:

a_n = a + (n - 1)d

Here, the $10^{th}$ term is given as 38:

a_{10} = a + (10 - 1)d

This implies:

38 = a + 9d

Let’s call this equation (i):

a + 9d = 38

Similarly, the $16^{th}$ term is 74:

a_{16} = a + (16 - 1)d

This gives us:

74 = a + 15d

Let’s call this equation (ii):

a + 15d = 74

Now, by subtracting equation (i) from equation (ii), we have:

a + 15d - (a + 9d) = 74 - 38

This simplifies to:

6d = 36

Thus, the common difference $d$ is:

d = 6

Now substitute $d = 6$ back into equation (i):

a + 9(6) = 38

This simplifies to:

a + 54 = 38

Solving for $a$ , we find:

a = -16

To find the $31^{st}$ term, $a_{31}$ :

a_{31} = a + (31 - 1)d

Substituting the known values:

a_{31} = -16 + 30(6)

= -16 + 180

= 164

Hence, the $31^{st}$ term of the A.P. is 164.

Question 17

Which term of the series :

21, 18, 15, …….. is -81 ?

Can any term of this series be zero ?

If yes, find the number of terms.

Answer:

Consider the arithmetic progression: 21, 18, 15, …

Here, the first term $a = 21$ and the common difference $d = 18 - 21 = -3$ .

To find which term is -81, let the $n^{\text{th}}$ term be -81.

$a_n = -81$
$a + (n - 1)d = -81$
$21 + (n - 1)(-3) = -81$
$21 - 3n + 3 = -81$
$24 - 3n = -81$
$3n = 24 + 81$
$3n = 105$
$n = 35$

Thus, the -81 is the 35th term of the series.

Now, to determine if any term in the series is zero, let the $r^{\text{th}}$ term be 0.

$a_r = 0$
$a + (r - 1)d = 0$
$21 + (r - 1)(-3) = 0$
$21 - 3r + 3 = 0$
$24 - 3r = 0$
$3r = 24$
$r = 8$

Hence, the 0 is the 8th term of this arithmetic progression.

Question 18

An A.P. consists of 60 terms. If the first and the last term be 7 and 125 respectively, find the 31^st term.

Answer:

We know that the first term of the arithmetic progression (A.P.) is given as $a = 7$ and the last term, which is the 60th term, is $a_{60} = 125$ .

The formula for the nth term of an A.P. is given by:
$a_n = a + (n - 1)d$
For the 60th term:
$a_{60} = a + (60 - 1)d$
Substituting the given values:
$125 = 7 + 59d$
Rearranging the equation, we have:
$125 - 7 = 59d$
$118 = 59d$
Solving for $d$ :
$d = \frac{118}{59} = 2$
Now, to find the 31st term, $a_{31}$ , we use:
$a_{31} = a + (31 - 1)d$
Substituting the known values:
$a_{31} = 7 + 30 \times 2$
$a_{31} = 7 + 60$
$a_{31} = 67$

Thus, the 31st term is 67.

Question 19

The sum of the 4^th term and the 8^th terms of an A.P. is 24 and the sum of 6^th and the 10^th terms of the same A.P. is 34. Find the first three terms of the A.P.

Answer:

Given:

The sum of the 4th and 8th terms of an arithmetic progression (A.P.) is 24:

a_4 + a_8 = 24

The sum of the 6th and 10th terms is 34:

a_6 + a_{10} = 34

Let’s solve the first equation:

$a + (4 - 1)d + a + (8 - 1)d = 24$
$a + 3d + a + 7d = 24$
$2a + 10d = 24$
$a + 5d = 12$

Label this as equation (iii).

Now, for the second equation:

$a + (6 - 1)d + a + (10 - 1)d = 34$
$a + 5d + a + 9d = 34$
$2a + 14d = 34$
$a + 7d = 17$

Label this as equation (iv).

Subtract equation (iii) from equation (iv):

$(a + 7d) - (a + 5d) = 17 - 12$
$2d = 5$
$d = \dfrac{5}{2}$

Substitute the value of $d$ back into equation (iv):

$a + 7 \times \dfrac{5}{2} = 17$
$a + \dfrac{35}{2} = 17$
$a = 17 - \dfrac{35}{2} = \dfrac{34 - 35}{2} = -\dfrac{1}{2}$

The first three terms of the A.P. are:

$-\dfrac{1}{2}, -\dfrac{1}{2} + \dfrac{5}{2}, -\dfrac{1}{2} + 2 \times \dfrac{5}{2}$
$= -\dfrac{1}{2}, \dfrac{-1 + 5}{2}, -\dfrac{1}{2} + 5$
$= -\dfrac{1}{2}, 2, \dfrac{9}{2}$

Hence, the first three terms of the A.P. are $-\dfrac{1}{2}, 2, \dfrac{9}{2}$ .

Question 20

If the third term of an A.P. is 5 and the seventh term is 9, find the 17^th term.

Answer:

The formula for the n^\text{th} term of an arithmetic progression (A.P.) is:

a_n = a + (n - 1)d

Here, the 3^\text{rd} term is given as 5, so:

a_3 = a + (3 - 1)d

This simplifies to:

5 = a + 2d

Thus, we have:

a + 2d = 5 \quad \text{...(i)}

Similarly, the 7^\text{th} term is given as 9:

a_7 = a + (7 - 1)d

This simplifies to:

9 = a + 6d

So, we have:

a + 6d = 9 \quad \text{...(ii)}

By subtracting equation (i) from equation (ii), we find:

a + 6d - (a + 2d) = 9 - 5

4d = 4

\Rightarrow d = 1

Substitute the value of $d$ back into equation (i):

a + 2(1) = 5

a + 2 = 5

\Rightarrow a = 3

Now, to find the 17^\text{th} term $a_{17}$ :

a_{17} = a + (17 - 1)d

= 3 + 16(1)

= 3 + 16

= 19

Hence, 17^\text{th} term of A.P. = 19.

Exercise 10(B)

Question 1(a)

Two A.P.’s have same common difference. If the difference between their 25^th terms is 8, the difference between their 50^th terms is :

(a) 16
(b) 5
(c) 8
(d) 25

Answer: (c) 8

Consider two arithmetic progressions (A.P.’s) with first terms $a_1$ and $a_2$ respectively, and a common difference $d$ .

The general formula for the $n$ -th term of an A.P. is:

a_n = a + (n - 1)d

For the 25th term of the first A.P., we have:

a_1 + 24d

Similarly, the 25th term of the second A.P. is:

a_2 + 24d

We are given that the difference between the 25th terms of these A.P.’s is 8:

(a_1 + 24d) - (a_2 + 24d) = 8

This simplifies to:

a_1 - a_2 = 8

Now, let’s find the difference between their 50th terms. The 50th term of the first A.P. is:

a_1 + 49d

And the 50th term of the second A.P. is:

a_2 + 49d

The difference between the 50th terms is:

(a_1 + 49d) - (a_2 + 49d) = a_1 - a_2

Since we have already determined that $a_1 - a_2 = 8$ , the difference between the 50th terms is also 8.

Hence, Option 3 is the correct option.

Question 1(b)

Ten times the 10^th term of an A.P. is equal to twenty times the 20^th term of the same A.P. The 30^th term of this A.P. is :

(a) 0
(b) 40
(c) 20
(d) 2 × (30 + 10)

Answer: (a) 0

Consider the first term of the arithmetic progression (A.P.) as $a$ and the common difference as $d$ .

The formula for the $n^{th}$ term of an A.P. is given by:

a_n = a + (n - 1)d

According to the problem statement, we have:

10a_{10} = 20a_{20}

Substituting the formula for the $10^{th}$ and $20^{th}$ terms, we get:

10[a + (10 - 1)d] = 20[a + (20 - 1)d]

Simplifying both sides:

10[a + 9d] = 20[a + 19d]

This leads to:

10a + 90d = 20a + 380d

Rearranging terms gives:

20a - 10a = 90d - 380d

10a = -290d

Solving for $a$ , we find:

a = -29d

Now, let’s find the $30^{th}$ term:

a_{30} = a + (30 - 1)d

= a + 29d

Substitute $a = -29d$ :

= -29d + 29d

= 0

Hence, Option 1 is the correct option.

Question 1(c)

The n^th term of an A.P. is 7n – 5. Its common difference is :

(a) 2
(b) 9
(c) 16
(d) 7

Answer: (d) 7

We’re given that the n $^{th}$ term of the arithmetic progression (A.P.) is expressed as 7n – 5.

To find the common difference, we first determine the (n – 1) $^{th}$ term:

(n - 1)\text{th term} = 7(n - 1) - 5 = 7n - 7 - 5 = 7n - 12.

The common difference $d$ in an A.P. is calculated by subtracting the (n – 1) $^{th}$ term from the n $^{th}$ term. Thus,

d = \text{n$^{th}$ term} - (n - 1)\text{th term}

Substituting the terms, we have:

d = (7n - 5) - (7n - 12)

Simplifying this expression gives:

d = 7n - 7n - 5 + 12 = 7.

∴ The common difference is 7. Hence, Option 4 is the correct option.

Question 1(d)

The 40^th term of an A.P. exceeds its 16^th term by 72. Then its common difference is :

(a) 40
(b) 40 – 16
(c) 72
(d) 3

Answer: (d) 3

Consider the first term of the arithmetic progression as $a$ and the common difference as $d$ .

The formula for the $n^{th}$ term of an A.P. is given by:

a_n = a + (n - 1)d

According to the problem, the $40^{th}$ term is 72 more than the $16^{th}$ term.

∴ $a_{40} - a_{16} = 72$

Substituting the formula for each term:

[a + (40 - 1)d] - [a + (16 - 1)d] = 72

This simplifies to:

a + 39d - [a + 15d] = 72

Notice that $a$ cancels out:

a - a + 39d - 15d = 72

Simplifying further, we get:

24d = 72

Dividing both sides by 24 gives:

d = \dfrac{72}{24} = 3

Thus, the common difference $d$ is 3.

Hence, Option 4 is the correct option.

Question 1(e)

The n^th term of the A.P. 6, 11, 16, 21, ……. is 106, the the value of n – 4 is :

(a) 17
(b) 15
(c) 16
(d) 20

Answer: (a) 17

Consider the arithmetic progression given: 6, 11, 16, 21, …….

Here, the first term $a$ is 6, and the common difference $d$ is calculated as $11 - 6 = 5$ .

We know the formula for the nth term of an A.P. is:

a_n = a + (n - 1) \times d

Given that the nth term $a_n$ is 106, substituting the values we have:

106 = 6 + (n - 1) \times 5

Simplifying, we get:

106 = 6 + 5n - 5

106 = 5n + 1

Subtracting 1 from both sides:

106 - 1 = 5n

5n = 105

Now, divide by 5:

n = \frac{105}{5} = 21

Thus, $n - 4$ becomes:

n - 4 = 21 - 4 = 17

Hence, Option 1 is the correct option.

Question 2

In an A.P. ten times of its tenth term is equal to thirty times of its 30^th term. Find its 40^th term.

Answer:

Consider the first term of the arithmetic progression (A.P.) to be $a$ and the common difference to be $d$ .

According to the given condition:

⇒ $10a_{10} = 30a_{30}$

⇒ $10[a + (10 - 1)d] = 30[a + (30 - 1)d]$

Simplifying the equation, we have:

⇒ $a + 9d = 3(a + 29d)$

Expanding and rearranging terms, we get:

⇒ $a + 9d = 3a + 87d$

⇒ $3a - a = 9d - 87d$

⇒ $2a = -78d$

⇒ $a = -39d$ .

Now, to find the 40th term $a_{40}$ :

∴ $a_{40} = a + (40 - 1)d$

Substituting the value of $a$ :

= $-39d + 39d$

= $0$ .

Hence, $a_{40} = 0$ .

Question 3

How many two-digit numbers are divisible by 3 ?

Answer:

Consider the sequence of two-digit numbers that can be divided by 3 without leaving a remainder: 12, 15, 18, …, 99. This sequence is an arithmetic progression (A.P.) where the first term $a = 12$ , the common difference $d = 15 - 12 = 3$ , and the last term is 99.

We need to determine the total number of terms, denoted by $n$ .

Using the formula for the $n^{th}$ term of an A.P., $a_n = a + (n - 1)d$ , we have:

99 = 12 + (n - 1) \times 3

Simplifying, we get:

99 = 12 + 3n - 3

99 = 9 + 3n

Subtract 9 from both sides:

99 - 9 = 3n

90 = 3n

Divide both sides by 3:

n = 30

Thus, there are 30 two-digit numbers that are divisible by 3.

Question 4

Which term of A.P. 5, 15, 25, ……. will be 130 more than its 31^st term ?

Answer:

In the given arithmetic progression (A.P.) 5, 15, 25, …, the first term $a$ is 5 and the common difference $d$ is calculated as $15 - 5 = 10$ .

We need to find the $n^{th}$ term that is 130 more than the 31st term.

∴ $a_n = 130 + a_{31}$

⇒ (a + (n – 1)d = 130 + a + (31 – 1)d)

⇒ (5 + (n – 1) \times 10 = 130 + 5 + 30 \times 10)

⇒ $5 + 10n - 10 = 135 + 300$

⇒ $10n - 5 = 435$

⇒ $10n = 440$

⇒ $n = 44$ .

Thus, the 44th term of the A.P. will be 130 more than its 31st term.

Question 5

Find the value of p, if x, 2x + p and 3x + 6 are in A.P.

Answer:

Given that x, 2x + p, and 3x + 6 are terms of an arithmetic progression, the difference between each consecutive pair of terms must be the same.

∴ We have:

(2x + p) - x = (3x + 6) - (2x + p)

This simplifies to:

x + p = x + 6 - p

Rearranging gives:

p + p = x + 6 - x

Thus, we find:

2p = 6

Solving for p, we get:

p = 3

Hence, p = 3.

Question 6

If the 3^rd and the 9^th terms of an arithmetic progression are 4 and -8 respectively, which term of it is zero ?

Answer:

Consider the formula for the $n^{th}$ term of an arithmetic progression (A.P.):

a_n = a + (n - 1)d

Here, we know that the $3^{rd}$ term is 4, so:

a_3 = a + 2d = 4 \quad \text{(equation 1)}

And the $9^{th}$ term is -8, so:

a_9 = a + 8d = -8 \quad \text{(equation 2)}

To find $d$ , subtract equation 1 from equation 2:

a + 8d - (a + 2d) = -8 - 4

This simplifies to:

6d = -12

Thus, $d = -2$ .

Substitute $d = -2$ back into equation 1:

a + 2(-2) = 4

This gives:

a - 4 = 4

So, $a = 8$ .

Now, to find which term is zero, set the $n^{th}$ term to zero:

a_n = a + (n - 1)d = 0

Substitute $a = 8$ and $d = -2$ :

8 + (n - 1)(-2) = 0

Simplifying, we get:

8 - 2n + 2 = 0

This leads to:

2n = 10

Therefore, $n = 5$ .

Hence, the $5^{th}$ term of the A.P. is zero.

Question 7

How many three digit numbers are divisible by 87?

Answer:

Consider the sequence of three-digit numbers that can be divided evenly by 87: 174, 261, …, 957. This sequence forms an arithmetic progression (A.P.) where the first term $a = 174$ , the common difference $d = 261 - 174 = 87$ , and the last term is 957.

Suppose the number of terms in this A.P. is $n$ .

∴ The last term $a_n = 957$

Using the formula for the $n^{th}$ term of an A.P., we have:

a + (n - 1)d = 957

Substituting the known values:

174 + (n - 1)(87) = 957

Simplifying further:

174 + 87n - 87 = 957

Combine like terms:

87n + 87 = 957

Subtract 87 from both sides:

87n = 957 - 87

Calculate the right-hand side:

87n = 870

Divide both sides by 87:

n = 10

Therefore, there are 10 three-digit numbers that are divisible by 87.

Question 8

For what value of n, the n^th term of A.P. 63, 65, 67, ……. and n^th term of A.P. 3, 10, 17, ….., are equal to each other?

Answer:

Consider the arithmetic progression (A.P.) 63, 65, 67, ……. Here, the first term $a$ is 63 and the common difference $d$ is calculated as $65 - 63 = 2$ .

Now, look at the A.P. 3, 10, 17, ……. For this sequence, the first term $a_1$ is 3, and the common difference $d_1$ is $10 - 3 = 7$ .

We know that the $n^{th}$ terms of both A.P.s are equal:

∴ $a + (n - 1)d = a_1 + (n - 1)d_1$

Substituting the values, we have:

⇒ $63 + (n - 1) imes 2 = 3 + (n - 1) imes 7$

Simplifying, we get:

⇒ $63 + 2n - 2 = 3 + 7n - 7$

⇒ $61 + 2n = 7n - 4$

Bringing like terms together, we have:

⇒ $7n - 2n = 61 + 4$

⇒ $5n = 65$

Solving for $n$ , we find:

⇒ $n = 13$ .

Hence, $n = 13$ .

Question 9

Determine the A.P. whose 3^rd term is 16 and the 7^th term exceeds the 5^th term by 12.

Answer:

Let’s start with the information that the 3rd term of the arithmetic progression (A.P.) is 16. This can be expressed as:

a + (3 - 1)d = 16

Simplifying, we have:

a + 2d = 16 \quad \text{...(i)}

Now, it’s given that the 7th term is 12 more than the 5th term. This gives us:

a_7 - a_5 = 12

Substituting the terms, we get:

[a + (7 - 1)d] - [a + (5 - 1)d] = 12

This simplifies to:

(a + 6d) - (a + 4d) = 12

Further simplification yields:

a - a + 6d - 4d = 12

2d = 12

Solving for $d$ , we find:

d = 6

Next, we substitute the value of $d$ back into equation (i):

a + 2(6) = 16

This simplifies to:

a + 12 = 16

a = 4

Thus, the A.P. is formed as follows:

a, (a + d), (a + 2d), \ldots

Substituting the values of $a$ and $d$ , we get:

4, (4 + 6), (4 + 2 \times 6), \ldots

Which results in:

4, 10, 16, \ldots

Hence, A.P. = 4, 10, 16, \ldots

Question 10

If numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n and its next two terms.

Answer:

Given that the numbers $n - 2$ , $4n - 1$ , and $5n + 2$ form an arithmetic progression (A.P.), the difference between consecutive terms must be the same.

∴ $(4n - 1) - (n - 2) = (5n + 2) - (4n - 1)$

⇒ $4n - n - 1 + 2 = 5n - 4n + 2 + 1$

⇒ $3n + 1 = n + 3$

⇒ $3n - n = 3 - 1$

⇒ $2n = 2$

⇒ $n = 1$ .

Substitute $n = 1$ back into the terms $n - 2$ , $4n - 1$ , and $5n + 2$ :

= $1 - 2$ , $4(1) - 1$ , $5(1) + 2$

= $-1$ , $3$ , $7$ .

This A.P. has a first term of $-1$ and a common difference of $3 - (-1) = 4$ .

The next two terms can be calculated as follows: $7 + 4 = 11$ and $7 + 2(4) = 15$ .

Hence, $n = 1$ and the next two terms of the A.P. are 11 and 15.

Question 11

Determine the value of k for which k^2 + 4k + 8, 2k^2 + 3k + 6 and 3k^2 + 4k + 4 are in A.P.

Answer:

Given that the expressions $k^2 + 4k + 8$ , $2k^2 + 3k + 6$ , and $3k^2 + 4k + 4$ form an arithmetic progression, the difference between consecutive terms must be the same.

∴ $2k^2 + 3k + 6 - (k^2 + 4k + 8) = 3k^2 + 4k + 4 - (2k^2 + 3k + 6)$

⇒ Simplifying both sides, we have:

2k^2 - k^2 + 3k - 4k + 6 - 8 = 3k^2 - 2k^2 + 4k - 3k + 4 - 6

⇒ This reduces to:

k^2 - k - 2 = k^2 + k - 2

⇒ Simplifying further, we find:

k^2 - k^2 + k + k = -2 + 2

⇒ $2k = 0$

⇒ Solving for $k$ , we get $k = 0$ .

Hence, $k = 0$ .

Question 12

State, true or false : if a, b and c are in A.P. then :

(i) 4a, 4b and 4c are in A.P.

(ii) a + 4, b + 4 and c + 4 are in A.P.

Answer:

(i) Consider that $a$ , $b$ , and $c$ form an arithmetic progression (A.P.).

Remember, when each term of an A.P. is multiplied by the same constant, the sequence remains an A.P.

Thus, multiplying each term by 4 results in $4a$ , $4b$ , and $4c$ , which are also in A.P.

Hence, yes the terms 4a, 4b and 4c are in A.P.

(ii) To determine if $a + 4$ , $b + 4$ , and $c + 4$ are in A.P., the difference between consecutive terms must be equal.

⇒ $(b + 4) - (a + 4) = (c + 4) - (b + 4)$

⇒ $b - a + 4 - 4 = c - b + 4 - 4$

⇒ $b - a = c - b$ [Since $a$ , $b$ , and $c$ form an A.P., this equation holds true]

Hence, yes the terms a + 4, b + 4 and c + 4 are in A.P.

Question 13

An A.P. consists of 57 terms of which 7^th term is 13 and the last term is 108. Find the 45^th term of this A.P.

Answer:

Consider the arithmetic progression (A.P.) with the first term as $a$ and the common difference as $d$ .

From the given information, we know:

a_{57} = 108

This implies:

a + (57 - 1)d = 108

a + 56d = 108 \quad \text{...(i)}

Additionally, it is given that:

a_{7} = 13

Thus:

a + (7 - 1)d = 13

a + 6d = 13 \quad \text{...(ii)}

Now, subtract equation (ii) from equation (i):

(a + 56d) - (a + 6d) = 108 - 13

a - a + 56d - 6d = 95

50d = 95

Solving for $d$ , we find:

d = 1.9

Substitute the value of $d$ back into equation (ii):

a + 6(1.9) = 13

a + 11.4 = 13

a = 13 - 11.4 = 1.6

To find the 45th term of the A.P., $a_{45}$ , calculate:

a_{45} = a + (45 - 1)d

= 1.6 + 44(1.9)

= 1.6 + 83.6

= 85.2

Hence, 45th term of A.P. = 85.2

Question 14

4^th term of an A.P. is equal to 3 times its first term and 7^th term exceeds twice the 3^rd term by 1. Find the first term and the common difference.

Answer:

Assume the first term of the arithmetic progression is $a$ and the common difference is $d$ .

From the problem statement, we have:

a_4 = 3a

This implies:

a + (4 - 1)d = 3a

Simplifying, we get:

a + 3d = 3a

Rearranging terms gives:

2a = 3d

Thus, the first term can be expressed as:

a = \dfrac{3d}{2} \quad \text{...(i)}

Additionally, it is given that the 7th term exceeds twice the 3rd term by 1:

a_7 - 2a_3 = 1

This translates to:

a + (7 - 1)d - 2[a + (3 - 1)d] = 1

Simplifying further:

a + 6d - 2(a + 2d) = 1

a + 6d - 2a - 4d = 1

a - 2a + 2d = 1

-a + 2d = 1

Now, substitute the expression for $a$ from (i) into this equation:

-\dfrac{3d}{2} + 2d = 1

Combine the terms:

\dfrac{-3d + 4d}{2} = 1

\dfrac{d}{2} = 1

Solving for $d$ gives:

d = 2

Substituting $d = 2$ back into equation (i), we find:

a = \dfrac{3 \times 2}{2} = 3

Therefore, the first term is 3 and the common difference is 2.

Question 15

The sum of the 2^nd term and the 7^th term of an A.P. is 30. If its 15^th term is 1 less than twice of its 8^th term, find the A.P.

Answer:

Consider the first term of the arithmetic progression (A.P.) as $a$ and the common difference as $d$ .

From the problem statement, we have:

a_2 + a_7 = 30

This expands to:

a + (2 - 1)d + a + (7 - 1)d = 30

Simplifying, we get:

a + d + a + 6d = 30

$2a + 7d = 30$ \hspace{1cm} \text{…(i)}

Additionally, it is given that:

2a_8 - 1 = a_{15}

This translates to:

2[a + (8 - 1)d] - 1 = a + (15 - 1)d

Simplifying further:

2[a + 7d] - 1 = a + 14d

2a + 14d - 1 = a + 14d

Rearranging terms, we find:

2a - a - 1 = 14d - 14d

a - 1 = 0

Thus, $a = 1$ .

Substituting $a = 1$ into equation (i):

2(1) + 7d = 30

2 + 7d = 30

7d = 28

$d = 4$ .

Therefore, the A.P. is given by:

a, (a + d), (a + 2d), \ldots

Substituting the values of $a$ and $d$ :

1, (1 + 4), (1 + 2.4), \ldots

1, 5, 9, \ldots

Hence, A.P. = 1, 5, 9, \ldots

Question 16

In an A.P. if m^th term is n and n^th term is m, show that its r^th term is (m + n – r).

Answer:

Consider the first term of the arithmetic progression as $a$ and the common difference as $d$ .

From the problem statement, we know:

a_m = n

This implies:

a + (m - 1)d = n

Simplifying, we get:

a + md - d = n

Thus, we can express $a$ as:

a = n - md + d \quad \text{...(i)}

Additionally, we have:

a_n = m

Which gives us:

a + (n - 1)d = m

Simplifying further:

a + nd - d = m

Substitute the value of $a$ from equation (i) into this equation:

n - md + d + nd - d = m

This simplifies to:

nd - md = m - n

Rearranging, we find:

d(n - m) = m - n

Thus, the common difference $d$ is:

d = \dfrac{m - n}{n - m} = \dfrac{m - n}{-(m - n)} = -1

Now, substituting the value of $d$ back into equation (i), we get:

a = n - m(-1) + (-1) = n + m - 1

The $r^{th}$ term, $a_r$ , is given by:

a_r = a + (r - 1)d

Substituting the known values:

a_r = n + m - 1 + (r - 1)(-1)

Simplifying gives:

a_r = n + m - 1 - r + 1

Finally, we have:

a_r = n + m - r

Hence, proved that $r^{th}$ term is $(m + n - r)$ .

Question 17

Which term of the A.P. 3, 10, 17, …….. will be 84 more than its 13^th term ?

Answer:

Consider the given arithmetic progression (A.P.) with first term $a = 3$ and common difference $d = 10 - 3 = 7$ .

To find the 13th term, we use the formula for the $n^{th}$ term of an A.P., which is $a_n = a + (n - 1)d$ . So for the 13th term:

a_{13} = 3 + (13 - 1) imes 7 = 3 + 12 imes 7 = 3 + 84 = 87.

Now, we need to determine which term is 84 more than the 13th term. Thus, we set up the equation:

a_n = 84 + a_{13}.

Substituting the value of $a_{13}$ , we have:

a + (n - 1)d = 84 + 87.

This simplifies to:

3 + (n - 1) imes 7 = 171.

Solving for $n$ , we get:

(n - 1) imes 7 = 168

n - 1 = 24

n = 25.

Therefore, the 25th term is 84 more than the 13th term.

Exercise 10(C)

Question 1(a)

The sum of 41 terms of an A.P. with middle term 40 is :

(a) 820
(b) 1640
(c) 2460
(d) none of these

Answer: (b) 1640

The number of terms in the arithmetic progression is 41.

The middle term is calculated as $\dfrac{41 + 1}{2} = \dfrac{42}{2} = 21$ , which means the 21st term is the middle term.

We know that the middle term is given as 40.

∴ $a_{21} = 40$

Applying the formula for the $n^{th}$ term of an A.P., we have:

a + (21 - 1)d = 40

⇒ $a + 20d = 40$ …………(1)

The formula for the sum of an A.P. is:

\dfrac{n}{2}[2a + (n - 1)d]

Substituting the given values, we get:

\dfrac{41}{2}[2a + (41 - 1)d]

= $\dfrac{41}{2}[2a + 40d]$

= $\dfrac{41}{2} \times 2 \times [a + 20d]$

Using equation (1), $a + 20d = 40$ , we substitute:

= $41 \times 40$

= 1640.

Therefore, the sum of the 41 terms of the A.P. is 1640.

Hence, Option 2 is the correct option.

Question 1(b)

The sum of all two digit numbers is :

(a) 9810
(b) 9045
(c) 4509
(d) 4905

Answer: (d) 4905

Consider the sequence of two-digit numbers: 10, 11, …, 99. This sequence forms an arithmetic progression (A.P.) where the first term $a = 10$ and the common difference $d = 11 - 10 = 1$ . The last term is 99.

To find the total number of terms $n$ , use the formula for the $n^{th}$ term of an A.P., (a_n = a + (n – 1)d).

99 = 10 + (n - 1) \times 1

Solve for $n$ :

\begin{aligned}99 = 10 + n - 1 \\99 = n + 9 \\n = 99 - 9 = 90\end{aligned}

Thus, there are 90 terms in this sequence.

To find the sum of these terms, apply the formula for the sum of $n$ terms of an A.P.:

\text{Sum} = \dfrac{n}{2}[2a + (n - 1)d]

Substitute the known values:

$\text{Sum} = \dfrac{90}{2}$ 2 \times 10 + (90 – 1) \times 1 $\begin{aligned} \\= 45[20 + 89] \\= 45 \times 109 \\= 4905\end{aligned}$

Hence, the sum of all two-digit numbers is 4905. Option 4 is the correct option.

Question 1(c)

The sum of A.P. 4, 7, 10, 13, …….. upto 20 terms is :

(a) 650
(b) 10 × 27
(c) 510
(d) 1300

Answer: (a) 650

Consider the arithmetic progression (A.P.) given by 4, 7, 10, 13, and so on, up to 20 terms.

Here, the first term $a$ is 4, and the common difference $d$ is calculated as $7 - 4 = 3$ .

To find the sum of the first 20 terms, use the formula for the sum of $n$ terms of an A.P.:

S_n = \dfrac{n}{2}[2a + (n - 1)d]

Applying this to our A.P., we have:

$S_{20} = \dfrac{20}{2}$ 2 \times 4 + (20 – 1) \times 3$$$$

Simplifying inside the brackets:

$= 10 \times$ 8 + 19 \times 3$$$$

= 10 \times [8 + 57]

= 10 \times 65

= 650

Therefore, the sum of the series up to 20 terms is 650.

Hence, Option 1 is the correct option.

Question 1(d)

The sum of 40 terms of the A.P. 7 + 10 + 13 + 16 + ….. is :

(a) 5240
(b) 2620
(c) 1310
(d) 2680

Answer: (b) 2620

Consider the arithmetic progression: 7, 10, 13, 16, and so on.

Here, the first term ((a)) is 7 and the common difference ((d)) is calculated as $10 - 7 = 3$ .

The formula for the sum of $n$ terms in an arithmetic progression is given by:

\text{Sum of } n \text{ terms} = \dfrac{n}{2}[2a + (n - 1)d]

Applying this formula to find the sum of the first 40 terms:

$= \dfrac{40}{2}$ 2 \times 7 + (40 – 1) \times 3 $\begin{aligned} \\= 20 \times \end{aligned}$ 14 + 39 \times 3 $\begin{aligned} \\= 20 \times [14 + 117] \\= 20 \times 131 \\= 2620.\end{aligned}$

Hence, Option 2 is the correct option.

Question 1(e)

The n^th term of an A.P. is 6n + 4. The sum of its first two terms is :

(a) 16
(b) 20
(c) 26
(d) none of these

Answer: (c) 26

We know the formula for the n-th term of the arithmetic progression is given as:

n^\text{th} \text{ term} = 6n + 4

To find the first term, substitute $n = 1$ :

\text{First term} = 6(1) + 4 = 10

Next, calculate the second term by substituting $n = 2$ :

\text{Second term} = 6(2) + 4 = 12 + 4 = 16

Now, find the sum of the first two terms:

\text{Sum of first two terms} = 10 + 16 = 26

Hence, Option 3 is the correct option.

Question 2

How many terms of the A.P. :

24, 21, 18, ……….. must be taken so that their sum is 78 ?

Answer:

In this Arithmetic Progression (A.P.), the first term $a$ is 24 and the common difference $d$ is -3.

We use the formula for the sum of the first $n$ terms of an A.P., which is:

S = \dfrac{n}{2}(2a + (n - 1)d)

We want this sum to be 78. Thus, we have:

78 = \dfrac{n}{2}(2 \times 24 + (n - 1)(-3))

Simplifying inside the brackets:

\Rightarrow 78 = \dfrac{n}{2}(48 - 3n + 3)

Multiply both sides by 2 to clear the fraction:

\Rightarrow 78 \times 2 = n(51 - 3n)

This leads to:

\Rightarrow 156 = 51n - 3n^2

Rearranging terms gives us a quadratic equation:

\Rightarrow 3n^2 - 51n + 156 = 0

Divide the entire equation by 3:

\Rightarrow n^2 - 17n + 52 = 0

To factorize, split the middle term:

\Rightarrow n^2 - 13n - 4n + 52 = 0

Factor by grouping:

\Rightarrow n(n - 13) - 4(n - 13) = 0

This results in:

\Rightarrow (n - 4)(n - 13) = 0

Setting each factor to zero gives:

\Rightarrow n - 4 = 0 \text{ or } n - 13 = 0

Thus, $n = 4$ or $n = 13$ .

Hence, no. of terms = 4 or 13.

Question 3

Find the sum of 28 terms of an A.P. whose n^th term is 8n – 5.

Answer:

We are given the n $^{\text{th}}$ term of an arithmetic progression (A.P.) as $a_n = 8n - 5$ .

First, find the first term $a_1$ :

a_1 = 8(1) - 5 = 3.

Next, find the 28 $^{\text{th}}$ term $a_{28}$ :

a_{28} = 8(28) - 5 = 219.

To calculate the sum of the first 28 terms, use the formula for the sum of an A.P., (S = \frac{n}{2}(a + l)), where $a$ is the first term and $l$ is the last term:

\begin{aligned}S &= \frac{28}{2}(3 + 219) \\&= 14 \times 222 \\&= 3108.\end{aligned}

Hence, sum = 3108.

Question 4(i)

Find the sum of all odd natural numbers less than 50.

Answer:

Consider the sequence of odd natural numbers below 50: 1, 3, 5, …, 49.

This sequence is an arithmetic progression (A.P.) where the first term $a = 1$ , the common difference $d = 2$ , and the last term $l = 49$ .

Suppose there are $n$ terms in this sequence.

∴ The $n^{th}$ term $a_n = 49$ .

Using the formula for the $n^{th}$ term of an A.P., we have:

⇒ $a + (n - 1)d = 49$

Substituting the known values:

⇒ $1 + 2(n - 1) = 49$

⇒ $1 + 2n - 2 = 49$

⇒ $2n - 1 = 49$

⇒ $2n = 50$

⇒ $n = 25$ .

Now, to find the sum of these terms, use the sum formula for an A.P.: $S = \dfrac{n}{2}(a + l)$ .

Thus,

S = \dfrac{25}{2}(1 + 49)

= \dfrac{25}{2} \times 50

$= 625$ .

Therefore, the sum of all odd natural numbers less than 50 is 625.

Question 4(ii)

Find the sum of first 12 natural numbers each of which is a multiple of 7.

Answer:

Consider the sequence of the first 12 natural numbers that are multiples of 7:

7, 14, 21, …, up to the 12th term.

This sequence forms an arithmetic progression (A.P.) where the first term $a = 7$ and the common difference $d = 7$ .

To find the sum of these terms, use the sum formula for an A.P.:

S = \dfrac{n}{2}(2a + (n - 1)d)

Substitute the given values into the formula:

S = \dfrac{12}{2}(2 \times 7 + (12 - 1) \times 7)

Simplify the expression:

= 6(14 + 77)

= 6 \times 91

= 546.

Therefore, the sum of the first 12 natural numbers that are multiples of 7 is 546.

Question 5

Find the sum of first 51 terms of an A.P. whose 2^nd and 3^rd terms are 14 and 18 respectively.

Answer:

We know that the second term of the A.P. is given by the formula:

a_2 = a + (2 - 1)d

Substituting the given value:

14 = a + d

This can be rearranged to:

$a = 14 - d$ ( \text{…(i)} )

Similarly, for the third term, we have:

a_3 = a + (3 - 1)d

Given that:

18 = a + 2d

Rearranging gives:

$a = 18 - 2d$ ( \text{…(ii)} )

Equating equations ( (i) ) and ( (ii) ), we find:

14 - d = 18 - 2d

Solving this, we have:

-d + 2d = 18 - 14

d = 4

Using the value of $d$ in equation ( (i) ):

a = 14 - 4 = 10

To find the sum of the first 51 terms, we use the formula for the sum of an A.P.:

S = \frac{n}{2}[2a + (n - 1)d]

Substituting the known values:

$\begin{aligned}S &= \frac{51}{2}$ 2 \times 10 + (51 – 1) \times 4 $\begin{aligned} \\&= \frac{51}{2} \times (20 + 200) \\&= \frac{51}{2} \times 220 \\&= 51 \times 110 \\&= 5610\end{aligned}\end{aligned}$

Thus, the sum of the first 51 terms is 5610.

Question 6

The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first n terms.

Answer:

Consider the first term of the arithmetic progression as $a$ and the common difference as $d$ .

We know the sum of the first 7 terms is 49.

\begin{aligned}\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\\Rightarrow 49 = \dfrac{7}{2}[2 \times a + (7 - 1)d] \\\Rightarrow 49 = \dfrac{7}{2}[2a + 6d] \\\Rightarrow 49 = 7(a + 3d) \\\Rightarrow a + 3d = 7 ........(i)\end{aligned}

Similarly, the sum of the first 17 terms is 289.

\begin{aligned}\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\\Rightarrow 289 = \dfrac{17}{2}[2 \times a + (17 - 1)d] \\\Rightarrow 289 = \dfrac{17}{2}[2a + 16d] \\\Rightarrow 289 = 17(a + 8d) \\\Rightarrow a + 8d = 17 ........(ii)\end{aligned}

Now, subtract equation (i) from equation (ii):

\begin{aligned}\Rightarrow a + 8d - (a + 3d) = 17 - 7 \\\Rightarrow a - a + 8d - 3d = 10 \\\Rightarrow 5d = 10 \\\Rightarrow d = 2.\end{aligned}

Substitute the value of $d$ back into equation (i):

\begin{aligned}\Rightarrow a + 3(2) = 7 \\\Rightarrow a + 6 = 7 \\\Rightarrow a = 1.\end{aligned}

Thus, the sum of the first $n$ terms is:

\begin{aligned}\text{Sum of n terms } = \dfrac{n}{2}[2a + (n - 1)d] \\= \dfrac{n}{2}[2(1) + (n - 1)2] \\= \dfrac{n}{2}[2 + 2n - 2] \\= \dfrac{n}{2} \times 2n \\= n^2.\end{aligned}

Therefore, the sum of the first $n$ terms is $n^2$ .

Question 7

The first term of an A.P. is 5, the last term is 45 and the sum of its term is 1000. Find the number of terms and the common difference of the A.P.

Answer:

We are given that the first term $a = 5$ , the last term $l = 45$ , and the sum of the terms $S = 1000$ .

Using the formula for the sum of an arithmetic progression:

S = \dfrac{n}{2}(a + l)

Substituting the known values:

1000 = \dfrac{n}{2}(5 + 45)

This simplifies to:

1000 = \dfrac{n}{2} \times 50

Solving for $n$ , we have:

n = \dfrac{1000 \times 2}{50}

n = 40

Now, to find the common difference $d$ , we know:

l = a_{40} = 45

This implies:

a + (40 - 1)d = 45

Substitute $a = 5$ :

5 + 39d = 45

Solving for $d$ :

39d = 40

d = \dfrac{40}{39}

Hence, $n = 40$ and $d = \dfrac{40}{39}$ .

Question 8

Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.

Answer:

To find the sum of natural numbers between 250 and 1000 that are divisible by 9, we start by calculating the smallest and largest multiples of 9 within this range.

Calculating, $\dfrac{250}{9}$ gives us $27\dfrac{7}{9}$ , indicating the smallest integer multiple is 28. Similarly, $\dfrac{1000}{9}$ equals $111\dfrac{1}{9}$ , so the largest integer multiple is 111.

Thus, the sequence of numbers divisible by 9 from 250 to 1000 is:

$28 \times 9, 29 \times 9, 30 \times 9, \ldots, 111 \times 9$ ,

which simplifies to:

252, 261, 270, \ldots, 999.

This sequence forms an arithmetic progression (A.P.) with a first term $a = 252$ , a common difference $d = 9$ , and a last term $l = 999$ .

To find the number of terms $n$ , use the formula for the $n$ -th term of an A.P., $a_n = a + (n - 1)d$ :

$999 = 252 + (n - 1) \times 9$ .

Simplifying,

$999 = 252 + 9n - 9$ ,

$999 = 9n + 243$ ,

$999 - 243 = 9n$ ,

$9n = 756$ ,

$n = 84$ .

Now, calculate the sum $S$ of this A.P. using the sum formula $S = \dfrac{n}{2}(a + l)$ :

$S = \dfrac{84}{2} \times (252 + 999)$ ,

$= 42 \times 1251$ ,

$= 52542$ .

Hence, sum = 52542.

Question 9

The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?

Answer:

We have the first term of the arithmetic progression (A.P.) as 34 and the last term as 700. The common difference is given as 18. To find the number of terms, denoted by $n$ , we use the formula for the $n$ -th term:

a_n = a + (n - 1) \times d

Substituting the known values:

700 = 34 + (n - 1) \times 18

Simplifying the equation:

\begin{aligned} 700 &= 34 + 18n - 18 \\ 700 &= 18n + 16 \\ 700 - 16 &= 18n \\ 684 &= 18n \\ n &= \dfrac{684}{18} = 38 \end{aligned}

Thus, there are 38 terms in the sequence.

Next, to find the sum of all terms in the A.P., we use the sum formula:

S = \dfrac{n}{2} \times (a + l)

Substituting the values we have:

\begin{aligned} S &= \dfrac{38}{2} \times (34 + 700) \\ S &= 19 \times 734 \\ S &= 13946 \end{aligned}

Therefore, the number of terms is 38 and their sum is 13946.

Question 10

In an A.P. the first term is 25, n^th term is -17 and the sum of n terms is 132. Find n and the common difference.

Answer:

We are given:

First term, $a = 25$
$n^{th}$ term, $a_n = -17$
Sum of $n$ terms, $S = 132$

From the formula for the $n^{th}$ term of an A.P., we have:

a + (n - 1)d = -17

Substituting the value of $a$ , we get:

25 + (n - 1)d = -17

This simplifies to:

(n - 1)d = -42

… (i)

Now, using the formula for the sum of $n$ terms in an A.P., we know:

S = \dfrac{n}{2}(2a + (n - 1)d)

Substitute the known values:

132 = \dfrac{n}{2}(2 \times 25 + (-42))

This simplifies to:

132 = \dfrac{n}{2}(50 - 42)

132 = \dfrac{8n}{2}

4n = 132

Solving for $n$ gives:

n = 33

Now, substitute $n = 33$ back into equation (i):

(33 - 1)d = -42

32d = -42

Solving for $d$ , we have:

d = -\dfrac{42}{32} = -\dfrac{21}{16}

Therefore, $n = 33$ and the common difference $d = -\dfrac{21}{16}$ .

Question 11(i)

If 18, a and (b – 3) are in A.P., then find the value of (2a – b).

Answer:

Consider the numbers 18, $a$ , and $(b - 3)$ forming an arithmetic progression (A.P.). In such a sequence, the difference between consecutive terms remains constant.

∴ We have:

a - 18 = (b - 3) - a

Simplifying this equation, we get:

a - 18 = b - 3 - a

Rearranging the terms gives:

a + a - 18 + 3 = b

This simplifies to:

2a - 15 = b

Thus, we can express $2a - b$ as follows:

2a - b = 15

Hence, the value of $2a - b = 15$ .

Question 11(ii)

Find the A.P. whose 4^th term is 9 and the sum of its 6^th term and 13^th term is 40.

Answer:

We know that the 4th term of the A.P. is 9. Therefore, we have:

a + (4 - 1)d = 9

This simplifies to:

a + 3d = 9

Solving for $a$ , we get:

a = 9 - 3d

…(1)

Next, we are told that the sum of the 6th term and the 13th term is 40. Hence:

a + (6 - 1)d + a + (13 - 1)d = 40

Simplifying, we have:

a + 5d + a + 12d = 40

which gives:

2a + 17d = 40

…(2)

Now, substitute the expression for $a$ from equation (1) into equation (2):

2(9 - 3d) + 17d = 40

Expanding and simplifying this, we find:

18 - 6d + 17d = 40

18 + 11d = 40

Subtracting 18 from both sides, we get:

11d = 22

Dividing by 11, we find:

d = 2

Substitute $d = 2$ back into equation (1) to find $a$ :

a = 9 - 3(2)

a = 9 - 6

a = 3

Thus, the arithmetic progression is:

3, 5, 7, 9, 11, ….

Hence, A.P is 3, 5, 7, 9, 11, …..

Question 12

The sum of n natural numbers is 5n^2 + 4n. Find its 8^th term.

Answer:

Consider that $T$ represents a term.

The sum of the first $n$ natural numbers is given by $S_n = 5n^2 + 4n$ .

For the first term ( $n = 1$ ), we have:

$S_1 = T_1 = 5(1)^2 + 4(1) = 5 + 4 = 9$ .

For the second term ( $n = 2$ ), calculate:

$S_2 = 5(2)^2 + 4(2) = 20 + 8 = 28$ .

To find the second term $T_2$ , subtract the sum of the first term from the sum of the first two terms:

$T_2 = S_2 - S_1 = 28 - 9 = 19$ .

Now, we know that:

T_2 = a + d

This implies:

19 = 9 + d

Therefore, $d = 10$ .

Now, for the eighth term $T_8$ , use the formula for the $n^{th}$ term of an arithmetic progression:

$T_8 = a + (8 - 1)d = 9 + 7(10) = 79$ .

Hence, the 8th term is 79.

Question 13

The fourth term of an A.P. is 11 and the eight term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

Answer:

Assume the first term of the arithmetic progression (A.P.) is $a$ and the common difference is $d$ .

Given that the fourth term is 11:

∴ $a_4 = 11$

This implies:

⇒ $a + (4 - 1)d = 11$

⇒ $a + 3d = 11$ …….(i)

Additionally, we know the eighth term exceeds twice the fourth term by 5:

∴ $a_8 = 2a_4 + 5$

This gives us:

⇒ $a + (8 - 1)d = 2[a + (4 - 1)d] + 5$

⇒ $a + 7d = 2a + 6d + 5$

Rearranging terms, we have:

⇒ $a - 2a + 7d - 6d = 5$

⇒ $-a + d = 5$ ……..(ii)

Now, let’s add equations (i) and (ii):

⇒ $a + 3d + (-a + d) = 11 + 5$

⇒ $4d = 16$

⇒ $d = 4$ .

Substituting $d = 4$ back into equation (i):

⇒ $a + 3(4) = 11$

⇒ $a + 12 = 11$

⇒ $a = -1$ .

Thus, the arithmetic progression is:

A.P. = $a, (a + d), (a + 2d), \, ...$

= $-1, 3, 7, \, ...$

To find the sum of the first 50 terms ( $S_{50}$ ):

\begin{aligned}S = \dfrac{n}{2}[2a + (n - 1)d] \\= \dfrac{50}{2}[2 \times (-1) + (50 - 1) \times 4] \\= 25[-2 + 196] \\= 25 \times 194 \\= 4850.\end{aligned}

Hence, A.P. = -1, 3, 7, ……….. and sum of first 50 terms = 4850.

Exercise 10(D)

Question 1(a)

k + 2, 2k + 7 and 4k + 12 are the first three terms of an A.P. The first term of this A.P. is :

(a) -2
(b) 0
(c) 2
(d) 3

Answer: (c) 2

The terms given are in an arithmetic progression: $k + 2$ , $2k + 7$ , and $4k + 12$ . For these to form an A.P., the difference between consecutive terms must be equal.

Thus, we have the equation:
$2k + 7 - (k + 2) = 4k + 12 - (2k + 7)$

Simplifying both sides, we get:
$(2k - k) + (7 - 2) = (4k - 2k) + (12 - 7)$

This simplifies to:
$k + 5 = 2k + 5$

Subtract $k$ and 5 from both sides:
$2k - k = 5 - 5$

This results in:
$k = 0$

Substituting $k = 0$ back into the first term $k + 2$ , we find:
$k + 2 = 0 + 2 = 2$

Hence, Option 3 is the correct option.

Question 1(b)

The sum of n terms of an A.P. is 3n^2. The second term of this A.P. is :

(a) 8
(b) 3
(c) 9
(d) 12

Answer: (c) 9

We are given that the sum of the first n terms of an arithmetic progression (A.P.) is expressed as 3n^2.

To find the sum of the first two terms, substitute n = 2 into the formula:

\text{Sum of 2 terms} = 3(2)^2 = 3 \times 4 = 12.

Similarly, for the sum of the first term, substitute n = 1:

\text{Sum of 1 term} = 3(1)^2 = 3 \times 1 = 3.

The second term of the A.P. is the difference between the sum of the first two terms and the sum of the first term:

\text{Second term} = 12 - 3 = 9.

Hence, Option 3 is the correct option.

Question 1(c)

If 5, 7 and 9 are in A.P. then which of the following is in A.P.?

(a) 5 × 7, 7 × 9 and 9 × 5
(b) 5 × 7, 7 × 7 and 9 × 7
(c) 2 × 5, 2 × 7 and 5 × 9
(d) 5 – 7, 7 – 9 and 9 – 5

Answer: (b) 5 × 7, 7 × 7 and 9 × 7

Consider the first sequence:

⇒ 5 \times 7, 7 \times 9, 9 \times 5

This gives us 35, 63, and 45.

Calculate the differences: 63 – 35 = 28 and 45 – 63 = -18.

The differences are not equal, ∴ this is not an A.P.

Now, examine the second sequence:

⇒ 5 \times 7, 7 \times 7, 9 \times 7

This results in 35, 49, and 63.

Check the differences: 49 – 35 = 14 and 63 – 49 = 14.

The differences are equal, ∴ this sequence is an A.P.

Hence, Option 2 is the correct option.

Question 2

Find three numbers in A.P. whose sum is 24 and whose product is 440.

Answer:

Consider three numbers in an arithmetic progression as ((a – d), a, (a + d)).

Given that their sum is 24:

a - d + a + a + d = 24

This simplifies to:

3a = 24

⇒ $a = 8$ .

For the product, we have:

(a - d)(a)(a + d) = 440

Substituting the value of $a$ :

(8 - d)(8)(8 + d) = 440

This can be rewritten as:

(8 - d)(8 + d) = \dfrac{440}{8}

⇒ $64 - d^2 = 55$

⇒ $d^2 = 64 - 55 = 9$

⇒ $d = \pm 3$

First, take $d = 3$ :

The sequence becomes ((8 – 3), 8, (8 + 3) = 5, 8, 11).

Alternatively, if $d = -3$ :

The sequence becomes ((8 – (-3)), 8, (8 + (-3)) = 11, 8, 5).

Therefore, the numbers in A.P. are 5, 8, 11 or 11, 8, 5.

Question 3

The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.

Answer:

Consider the angles of the quadrilateral as follows:

$a$ , $a + d$ , $a + 2d$ , $a + 3d$ .

Since the sum of angles in any quadrilateral is $360^\circ$ , we have:

a + (a + d) + (a + 2d) + (a + 3d) = 360^\circ

Simplifying, we get:

4a + 6d = 360^\circ

Now, divide the entire equation by 2:

2(2a + 3d) = 360^\circ

This simplifies to:

2a + 3d = 180^\circ

Given that the common difference $d = 20^\circ$ , substitute this into the equation:

2a + 3(20) = 180^\circ

2a + 60 = 180^\circ

Subtract $60$ from both sides:

2a = 180^\circ - 60 = 120^\circ

Divide by 2 to find $a$ :

$a = 60^\circ$ .

Hence, angles = 60°, 80°, 100°, 120°.

Question 4

Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.

Answer:

Consider the four terms of an arithmetic progression as ((a – 3d), (a – d), (a + d), (a + 3d)).

The sum of these four terms is given by:

a - 3d + a - d + a + d + a + 3d = 96

This simplifies to:

4a = 96

\Rightarrow a = 24

Thus, the terms become:

$24 - 3d, 24 - d, 24 + d, 24 + 3d$ .

We know the ratio of the product of the means to the product of the extremes is 15:7. Therefore, we have:

\dfrac{(24 - d)(24 + d)}{(24 - 3d)(24 + 3d)} = \dfrac{15}{7}

Simplifying the equation, we get:

\dfrac{576 - d^2}{576 - 9d^2} = \dfrac{15}{7}

Cross-multiplying gives:

7(576 - d^2) = 15(576 - 9d^2)

\Rightarrow 4032 - 7d^2 = 8640 - 135d^2

\Rightarrow 135d^2 - 7d^2 = 8640 - 4032

\Rightarrow 128d^2 = 4608

\Rightarrow d^2 = 36

\Rightarrow d = \pm 6

If $d = 6$ , the terms are:

((24 – 3 \times 6), (24 – 6), (24 + 6), (24 + 3 \times 6) = 6, 18, 30, 42).

If $d = -6$ , the terms are:

((24 – 3 \times (-6)), (24 – (-6)), (24 + (-6)), (24 + 3 \times (-6)) = 42, 30, 18, 6).

Hence, four parts of 96 are 6, 18, 30, 42 or 42, 30, 18, 6.

Question 5

Find five numbers in A.P. whose sum is $12\dfrac{1}{2}$ and the ratio of the first to the last term is 2 : 3.

Answer:

Consider the sequence of numbers in an arithmetic progression as ((a – 2d), (a – d), a, (a + d), (a + 2d)).

We know the total sum of these numbers is given as $12\dfrac{1}{2}$ .

a - 2d + a - d + a + a + d + a + 2d = 12\dfrac{1}{2}

Simplifying, we have:

\Rightarrow 5a = \dfrac{25}{2}

Solving for $a$ , we find:

\Rightarrow a = \dfrac{5}{2} = 2.5

It is also given that the ratio of the first term to the last term is $2 : 3$ .

\dfrac{a - 2d}{a + 2d} = \dfrac{2}{3}

Cross-multiplying gives:

\Rightarrow 3(a - 2d) = 2(a + 2d)

Expanding both sides, we get:

\Rightarrow 3a - 6d = 2a + 4d

Rearranging terms, we have:

\Rightarrow 3a - 2a = 4d + 6d

This simplifies to:

\Rightarrow a = 10d

Substituting $a = 2.5$ into the equation, we find:

\Rightarrow \dfrac{5}{2} = 10d

Solving for $d$ , we have:

\Rightarrow d = \dfrac{5}{2 \times 10} = \dfrac{1}{4} = 0.25

Thus, the sequence of numbers is:

2.5 - 2 \times 0.25, 2.5 - 0.25, 2.5, 2.5 + 0.25, 2.5 + 2 \times 0.25

Calculating these values, we get:

2, 2.25, 2.5, 2.75, 3

Hence, the numbers are 2, 2.25, 2.5, 2.75, 3.

Question 6

Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.

Answer:

Consider the three numbers in arithmetic progression as $a - d$ , $a$ , and $a + d$ .

According to the problem, the sum of these numbers is 207:

(a - d) + a + (a + d) = 207

This simplifies to:

3a = 207

Solving for $a$ , we find:

a = \frac{207}{3} = 69

It’s given that the product of the two smaller numbers is 4623:

(a - d) \times a = 4623

Substituting $a = 69$ into the equation:

69 \times (69 - d) = 4623

Expanding and simplifying gives:

4761 - 69d = 4623

Rearranging, we have:

69d = 4761 - 4623

69d = 138

Solving for $d$ , we get:

d = \frac{138}{69} = 2

Thus, the numbers in the sequence are:

69 - 2, 69, 69 + 2

This results in the numbers 67, 69, and 71. Therefore, the parts can be either 67, 69, 71 or in reverse order: 71, 69, 67.

Question 7

The sum of three numbers in A.P. is 15 and the sum of the squares of the extreme terms is 58. Find the numbers.

Answer:

Consider the three numbers in arithmetic progression as $a - d$ , $a$ , and $a + d$ .

According to the problem, the sum of these numbers is 15:

⇒ $(a - d) + a + (a + d) = 15$

This simplifies to:

⇒ $3a = 15$

⇒ $a = 5$ .

We are also given that the sum of the squares of the extreme terms is 58:

⇒ $(a - d)^2 + (a + d)^2 = 58$

Substituting $a = 5$ , we have:

⇒ $(5 - d)^2 + (5 + d)^2 = 58$

Expanding both squares:

⇒ $25 + d^2 - 10d + 25 + d^2 + 10d = 58$

Simplifying gives:

⇒ $50 + 2d^2 = 58$

⇒ $2d^2 = 8$

⇒ $d^2 = 4$

⇒ $d = \pm2$

First, consider $d = 2$ :

Numbers are $(5 - 2)$ , $5$ , $(5 + 2)$ , which are $3$ , $5$ , $7$ .

Now, consider $d = -2$ :

Numbers are $(5 - (-2))$ , $5$ , $(5 + (-2))$ , which are $7$ , $5$ , $3$ .

Hence, numbers = 3, 5, 7 or 7, 5, 3.

Question 8

Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Answer:

Consider the four numbers in an arithmetic progression as $a - 3d$ , $a - d$ , $a + d$ , and $a + 3d$ .

We know their sum is 20:

a - 3d + a - d + a + d + a + 3d = 20

Simplifying, we find:

4a = 20

a = 5

Next, we’re given that the sum of their squares is 120:

(a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 120

Substituting $a = 5$ , we have:

(5 - 3d)^2 + (5 - d)^2 + (5 + d)^2 + (5 + 3d)^2 = 120

Expanding each term, we get:

25 + 9d^2 - 30d + 25 + d^2 - 10d + 25 + d^2 + 10d + 25 + 9d^2 + 30d = 120

Combining like terms results in:

100 + 20d^2 = 120

Subtracting 100 from both sides gives:

20d^2 = 20

Dividing by 20, we find:

d^2 = 1

Thus, $d = \pm 1$ .

If $d = 1$ :

The numbers are $(5 - 3 \times 1)$ , $(5 - 1)$ , $(5 + 1)$ , $(5 + 3 \times 1)$ , which simplifies to 2, 4, 6, 8.

If $d = -1$ :

The numbers are $(5 - 3 \times (-1))$ , $(5 - (-1))$ , $(5 + (-1))$ , $(5 + 3 \times (-1))$ , simplifying to 8, 6, 4, 2.

Hence, numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

Question 9

Insert one arithmetic mean between 3 and 13.

Answer:

To find the arithmetic mean between the numbers 3 and 13, calculate the average of these two numbers. This is done by adding them together and dividing by 2. So, the arithmetic mean is given by $\dfrac{3 + 13}{2} = \dfrac{16}{2}$ , which simplifies to 8.

Hence, arithmetic mean between 3 and 13 = 8.

Question 10

The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.

Answer:

We have a sequence of angles in an arithmetic progression (A.P.) where the smallest angle is 120° and the common difference is 5°.

Let’s denote the number of sides of the polygon as $n$ . The sum of the interior angles of a polygon with $n$ sides is given by the formula $(2n - 4) \times 90°$ .

The sum of the angles in the A.P. can be calculated using the formula for the sum of an arithmetic sequence:

$\frac{n}{2}$ 2 \times 120° + (n – 1) \times 5°$$$$

Setting the sum of the A.P. equal to the sum of the interior angles, we have:

$\begin{aligned} \frac{n}{2}$ 2 \times 120° + (n – 1) \times 5° $\begin{aligned} &= (2n - 4) \times 90° \\ \Rightarrow \frac{n}{2}[240° + 5°n - 5°] &= 180°n - 360° \\ \Rightarrow 240°n + 5°n^2 - 5°n &= 360°n - 720° \\ \Rightarrow 5°n^2 + 235°n &= 360°n - 720° \\ \Rightarrow 5°n^2 - 125°n + 720° &= 0 \\ \Rightarrow n^2 - 25n + 144 &= 0 \end{aligned}\end{aligned}$

We can factorize this quadratic equation as follows:

\begin{aligned} n^2 - 16n - 9n + 144 &= 0 \\ \Rightarrow n(n - 16) - 9(n - 16) &= 0 \\ \Rightarrow (n - 9)(n - 16) &= 0 \end{aligned}

Thus, the solutions for $n$ are $n = 9$ or $n = 16$ .

Hence, the number of sides of the polygon can be 9 or 16.

Question 11

If p^th term of an A.P. is q and its q^th term is p, show that its (p + q)^th term is zero.

Answer:

Consider the first term of the arithmetic progression (A.P.) as $a$ and the common difference as $d$ .

For the $p^{th}$ term, we have:

T_p = a + (p - 1)d = q \quad \text{...(1)}

And for the $q^{th}$ term:

T_q = a + (q - 1)d = p \quad \text{...(2)}

By subtracting equation (2) from equation (1), we find:

a + (p - 1)d - [a + (q - 1)d] = q - p

This simplifies to:

(p - 1)d - (q - 1)d = q - p

Rearranging gives:

pd - d - qd + d = q - p

pd - qd = q - p

(p - q)d = q - p

(p - q)d = -(p - q)

Dividing both sides by $(p - q)$ , we get:

d = \dfrac{-(p - q)}{(p - q)}

d = -1

Substitute $d = -1$ back into equation (1):

a + (p - 1)(-1) = q

a - p + 1 = q

a = p + q - 1

Now, for the $(p + q)^{th}$ term:

T_{p + q} = a + (p + q - 1)d

Substituting $a = p + q - 1$ and $d = -1$ :

(p + q - 1) + (p + q - 1)(-1)

= (p + q - 1) - (p + q - 1)

= 0

Therefore, the $(p + q)^{th}$ term of the A.P. is zero.

Question 12

If a, b and c are p^th, q^th and r^th terms of an A.P., prove that

a(q – r) + b(r − p) + c(p − q) = 0

Answer:

Consider the first term of the arithmetic progression as $t$ and the common difference as $d$ .

The formula for the $n$ th term of an A.P. is $a_n = t + (n - 1) d$ .

For the $p$ th term, we have:

a = t + (p - 1)d \quad \text{...(1)}

For the $q$ th term, we have:

b = t + (q - 1)d \quad \text{...(2)}

For the $r$ th term, we have:

c = t + (r - 1)d \quad \text{...(3)}

Subtract equation (2) from equation (1):

$t + (p - 1)d - \left$ t + (q – 1)d\right $= a - b$

Simplifying, we find:

(p - 1 - q + 1)d = a - b

(p - q)d = a - b

Thus, the common difference $d$ can be expressed as:

d = \dfrac{(a - b)}{(p - q)} \quad \text{...(4)}

Next, subtract equation (3) from equation (2):

$t + (q - 1)d - \left$ t + (r – 1)d\right $= b - c$

Simplifying, we obtain:

(q - 1 - r + 1)d = b - c

(q - r)d = b - c

Thus, the common difference $d$ can also be expressed as:

d = \dfrac{b - c}{q - r} \quad \text{...(5)}

From equations (4) and (5), we equate the two expressions for $d$ :

\dfrac{(a - b)}{(p - q)} = \dfrac{b - c}{q - r}

Cross-multiplying gives:

(a - b)(q - r) = (b - c)(p - q)

Expanding both sides, we have:

aq - ar - bq + br = bp - bq - cp + cq

Rearranging terms, we get:

bp - cp + cq - aq + ar - br - bq + bq = 0

Grouping terms, it becomes:

(-aq + ar) + (bp - br) + (-cp + cq) = 0

This simplifies to:

-a(q - r) - b(r - p) - c(p - q) = 0

Finally, we rewrite it as:

a(q - r) + b(r - p) + c(p - q) = 0

Thus, it is proved that $a(q - r) + b(r - p) + c(p - q) = 0$ .

Question 13

Show that a^2, b^2 and c^2 are in A.P., if

$\dfrac{1}{b + c}, \dfrac{1}{c + a} \text{ and } \dfrac{1}{a + b}$ are in A.P.

Answer:

Given that $\dfrac{1}{b + c}, \dfrac{1}{c + a}, \dfrac{1}{a + b}$ form an arithmetic progression, we can state:

\dfrac{1}{c + a} - \dfrac{1}{b + c} = \dfrac{1}{a + b} - \dfrac{1}{c + a}

This implies:

\dfrac{(b + c) - (c + a)}{(c + a)(b + c)} = \dfrac{(c + a) - (a + b)}{(a + b)(c + a)}

Simplifying the numerators:

\dfrac{b - a}{(c + a)(b + c)} = \dfrac{c - b}{(a + b)(c + a)}

Cross-multiplying gives:

\dfrac{b - a}{b + c} = \dfrac{c - b}{a + b}

Expanding the terms, we have:

(b - a)(a + b) = (c - b)(b + c)

Distributing terms results in:

ab + b^2 - a^2 - ab = bc + c^2 - b^2 - bc

Simplifying further:

b^2 - a^2 = c^2 - b^2

This shows:

b^2 - a^2 = c^2 - b^2

The equal differences between consecutive terms confirm that:

a^2, b^2, c^2 are in A.P.

Question 14

A man saved ₹ 7,65,000 in 10 years. In each year, after the first, he saved ₹ 6,000 more than he did in the preceding year. How much did he save in the first seven years?

Answer:

The total savings over 10 years is ₹ 7,65,000. Each subsequent year, the man saved ₹ 6,000 more than the previous year, so the common difference, $d$ , is 6000.

Assume his savings in the first year was ₹ $a$ . The number of years, $n$ , is 10.

The formula for the sum of an arithmetic progression is:

S_n = \dfrac{n}{2}[2a + (n - 1)d]

Substitute the known values into the formula:

\begin{aligned}\Rightarrow 765000 = \dfrac{10}{2}[2a + 9(6000)] \\\Rightarrow 765000 = 5[2a + 54000] \\\Rightarrow 765000 = 10a + 270000 \\\Rightarrow 10a = 765000 - 270000 \\\Rightarrow 10a = 495000 \\\Rightarrow a = \dfrac{495000}{10} \\\Rightarrow a = 49500\end{aligned}

Now, calculate the savings for the first seven years:

\begin{aligned}\Rightarrow S_7 = \dfrac{7}{2}[2(49500) + (7 - 1)(6000)] \\= 3.5[99000 + 6(6000)] \\= 3.5(99000 + 36000) \\= 3.5(135000) \\= ₹ 4,72,500.\end{aligned}

Therefore, the amount saved in the first 7 years is ₹ 4,72,500.

Question 15

If S~n denotes the sum of first n terms of an A.P., prove that :

S~12 = 3(S~8 − S~4)

Answer:

To find the sum of the first $n$ terms in an arithmetic progression (A.P.), we use the formula:

S_n = \dfrac{n}{2}[2a + (n - 1)d]

where $a$ is the first term and $d$ is the common difference.

Let’s calculate $S_{12}$ :

$S_{12} = \dfrac{12}{2}[2a + 11d]$
$\Rightarrow S_{12} = 6(2a + 11d)$
$\Rightarrow S_{12} = 12a + 66d \quad \text{....(1)}$

Now, calculate $S_{8}$ :

$S_{8} = \dfrac{8}{2}[2a + 7d]$
$\Rightarrow S_{8} = 4(2a + 7d)$
$\Rightarrow S_{8} = 8a + 28d \quad \text{....(2)}$

Next, calculate $S_{4}$ :

$S_{4} = \dfrac{4}{2}[2a + 3d]$
$\Rightarrow S_{4} = 2(2a + 3d)$
$\Rightarrow S_{4} = 4a + 6d \quad \text{....(3)}$

Subtract equation (3) from equation (2):

$S_{8} - S_{4} = (8a + 28d) - (4a + 6d)$
$\Rightarrow S_{8} - S_{4} = 8a + 28d - 4a - 6d$
$\Rightarrow S_{8} - S_{4} = 4a + 22d$

Multiply this result by 3:

$3(S_{8} - S_{4}) = 3(4a + 22d)$
$\Rightarrow 3(S_{8} - S_{4}) = 12a + 66d$

Notice that:

3(S_{8} - S_{4}) = S_{12}

Hence, proved that ( S_{12} = 3(S_{8} − S_{4}) ).

Question 16

18^th term of an A.P. is equal to 4 times its 4^th term and the 6^th term exceeds twice the 2^nd term by 4. Find the sum of the first 9 terms of this A.P.

Answer:

In an arithmetic progression, the formula for the $n^{th}$ term is given by:

$a_n = a + (n - 1)d$ .

Here, it’s mentioned that the $18^{th}$ term is 4 times the $4^{th}$ term. This implies:

a_{18} = a + 17d

a~4 = a + 3d

⇒ $a + 17d = 4(a + 3d)$

⇒ $a + 17d = 4a + 12d$

⇒ $a - 4a + 17d - 12d = 0$

⇒ $-3a + 5d = 0$

⇒ $5d = 3a$

⇒ $a = \dfrac{5d}{3}$ \hspace{1em}…(1)

Another condition given is that the $6^{th}$ term exceeds twice the $2^{nd}$ term by 4. So,

a~6 = a + 5d

a~2 = a + d

⇒ $a + 5d = 2(a + d) + 4$

⇒ $a + 5d = 2a + 2d + 4$

⇒ $a + 5d - 2a - 2d = 4$

⇒ $3d - a = 4$

⇒ $3d - a = 4$ \hspace{1em}…(2)

Now, substitute the value of $a$ from equation (1) into equation (2):

⇒ $3d - \dfrac{5d}{3} = 4$

⇒ $\dfrac{9d - 5d}{3} = 4$

⇒ $4d = 12$

⇒ $d = \dfrac{12}{4}$

⇒ $d = 3$

Substitute $d = 3$ back into equation (1):

⇒ $a = \dfrac{5d}{3}$

⇒ $a = \dfrac{5(3)}{3}$

⇒ $a = 5$

To find the sum of the first 9 terms, we use the sum formula for an arithmetic progression:

S_n = \dfrac{n}{2}[2a + (n - 1)d]

where $a$ is the first term and $d$ is the common difference.

Thus, $\begin{aligned}S_{9} = \dfrac{9}{2}[2(5) + 8(3)] \\= \dfrac{9}{2}[10 + 24] \\= \dfrac{9}{2}[34] \\= 9 \times 17 \\= 153.\end{aligned}$

Hence, the sum of the first 9 terms of this A.P. = 153.

Test Yourself

Question 1(a)

If A, B and C are three arithmetic progressions (APs) as given below :

A = 2, 4, 6, 8, …….. upto n terms

B = 3, 6, 9, 12, …… upto n terms

C = 0, 4, 8, 12, …… upto n terms, then

out of A + B, A – C, C – B and B – A which is/are A.P. ?

(a) A + B
(b) A – C
(c) C – B
(d) All are A.P.

Answer: (d) All are A.P.

We have three arithmetic progressions given:

A: 2, 4, 6, 8, … up to n terms
B: 3, 6, 9, 12, … up to n terms
C: 0, 4, 8, 12, … up to n terms

Let’s examine the sequences formed by combining these APs:

A + B:
– The sequence becomes: 2 + 3, 4 + 6, 6 + 9, 8 + 12, …
– This simplifies to: 5, 10, 15, 20, …
– Notice that this sequence is an arithmetic progression with a common difference of 5.

A – C:
– The sequence is: 2 – 0, 4 – 4, 6 – 8, 8 – 12, …
– Simplifying gives: 2, 0, -2, -4, …
– This is also an arithmetic progression with a common difference of -2.

C – B:
– The sequence is: 0 – 3, 4 – 6, 8 – 9, 12 – 12, …
– Simplifying results in: -3, -2, -1, 0, …
– This sequence forms an arithmetic progression with a common difference of 1.

∴ All the sequences A + B, A – C, and C – B are arithmetic progressions. Hence, Option 4 is the correct option.

Question 1(b)

In an A.P. a = -36, d = 18 and l = 36, then n is :

(a) 10
(b) 5
(c) 15
(d) 20

Answer: (b) 5

We’re given the first term $a = -36$ , the common difference $d = 18$ , and the last term $l = 36$ of an arithmetic progression. We need to determine the number of terms $n$ .

Recall the formula for the $n$ th term of an arithmetic progression:

a_n = a + (n - 1) imes d

Here, the last term $a_n$ is 36. So, substituting the known values into the formula:

36 = -36 + (n - 1) imes 18

Simplifying the equation:

36 = -36 + 18n - 18

Adding 36 and 18 to both sides, we have:

36 + 36 + 18 = 18n

Thus, we find:

18n = 90

Solving for $n$ :

n = \frac{90}{18} = 5

Hence, Option 2 is the correct option.

Question 1(c)

Do the numbers 1^2, 5^2, 7^2, 73 ….. form an A.P. ? If yes, its next term will be :

(a) Yes, 11^2
(b) No
(c) Yes, 97
(d) Yes, 24

Answer: (c) Yes, 97

Consider the sequence given:

1^2, 5^2, 7^2, 73, …

which simplifies to:

1, 25, 49, 73, …

To determine if these numbers form an arithmetic progression (A.P.), we need to check if the difference between consecutive terms is constant. Calculate the differences:

The difference between the second term and the first term is 25 – 1 = 24.
The difference between the third term and the second term is 49 – 25 = 24.

Notice that both differences are the same, 24, indicating that the sequence has a common difference.

∴ The sequence is an A.P.

To find the next term, add the common difference to the last term:

Next term = 73 + 24 = 97.

Option 3 is the correct option.

Question 1(d)

The sum of first 10 even natural numbers is :

(a) 120
(b) 110
(c) 65
(d) 120

Answer: (b) 110

Consider the sequence of the first 10 even natural numbers: 2, 4, 6, 8, 10, and so on, up to 10 terms.

This sequence forms an arithmetic progression (A.P.) where:

The first term (a) is 2.
The common difference (d) is 2.

To find the sum of these terms, use the formula for the sum of n terms of an A.P.:

\text{Sum of } n \text{ terms} = \dfrac{n}{2} [2a + (n - 1)d]

Substitute the known values into the formula:

$= \dfrac{10}{2}$ 2 \times 2 + (10 – 1) \times 2 $\begin{aligned} \\= 5 \times \end{aligned}$ 4 + 9 \times 2 $\begin{aligned} \\= 5 \times [4 + 18] \\= 5 \times 22 \\= 110.\end{aligned}$

Hence, Option 2 is the correct option.

Question 1(e)

For the given numbers $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}............$

Assertion (A):

To find whether these terms form an A.P. or not. Express each term as the product of a natural number and $\sqrt{3 }$ i.e.,

$\sqrt{3} = 1 \times \sqrt{3}, \sqrt{12} = 2\sqrt{3}, \sqrt{27} = 3\sqrt{3}, \sqrt{48} = 4\sqrt{3}$ , etc.

Reason (R): Since, for the given number difference between the consecutive term is same. It is an A.P.

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

Consider the sequence given:

\Rightarrow \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}............

Express each term as a product of a natural number and $\sqrt{3}$ :

\Rightarrow \sqrt{3}, \sqrt{4 \times 3}, \sqrt{9 \times 3}, \sqrt{16 \times 3}............

Simplifying further:

\Rightarrow \sqrt{3}, \sqrt{4} \times \sqrt{3}, \sqrt{9} \times \sqrt{3}, \sqrt{16} \times \sqrt{3}............

This becomes:

\Rightarrow \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}............

Now, calculate the difference between consecutive terms:

The difference between the first and second terms is $2\sqrt{3} - \sqrt{3} = \sqrt{3}$ .
The difference between the second and third terms is $3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$ .

Thus, the first term is $\sqrt{3}$ and the common difference is $\sqrt{3}$ .

∴ The assertion is true.

Since the difference between consecutive terms is consistent, the sequence is indeed an arithmetic progression. Therefore, the reason is also true and it correctly supports the assertion.

Hence, option 3 is the correct option.

Question 1(f)

5, 8, 11, 14, …………… are in AP.

Assertion (A): $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,$ …………… are also in AP.

Reason (R): If each term of a given A.P. is divided by the same non zero number, the resulting sequence is an A.P..

(a) A is true, R is false.
(b) A is false, R is true.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (c) Both A and R are true and R is correct reason for A.

The sequence 5, 8, 11, 14, … is an arithmetic progression (AP) where the first term is 5 and the common difference is calculated as follows: 8 – 5 = 3, 11 – 8 = 3. Thus, the common difference is 3.

Consider the new sequence: $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7, ...$ . This sequence is derived by dividing each term of the original sequence by 2.

Let’s verify if this new sequence forms an AP:

The difference between the second and first term is $4 - \dfrac{5}{2} = \dfrac{8 - 5}{2} = \dfrac{3}{2}$ .
The difference between the third and second term is $\dfrac{11}{2} - 4 = \dfrac{11 - 8}{2} = \dfrac{3}{2}$ .
The difference between the fourth and third term is $7 - \dfrac{11}{2} = \dfrac{14 - 11}{2} = \dfrac{3}{2}$ .

Notice that the common difference in this sequence is $\dfrac{3}{2}$ , which is consistent across the terms, indicating that this sequence is indeed an AP.

Therefore, the assertion is true.

The reason states that if every term of an AP is divided by the same non-zero number, the resulting sequence is also an AP. This is indeed correct, as demonstrated by our calculations.

Hence, option 3 is the correct option.

Question 1(g)

An A.P. with 3^rd term = -8 and 9^th term = 4.

Assertion (A): Common difference = -2.

Reason (R): If first term of the A.P. is a, then (a + 8d) – (a + 2d) = -8 – 4.

(a) A is true, R is false.
(b) Both A and R are false.
(c) Both A and R are true and R is correct reason for A.
(d) Both A and R are true and R is incorrect reason for A.

Answer: (b) Both A and R are false.

In the given arithmetic progression (A.P.), the 3rd term is -8 and the 9th term is 4.

Let the first term be $a$ and the common difference be $d$ .

The general term formula for an A.P. is:

a_n = a + (n - 1)d

For the 3rd term:

$a_3 = a + (3 - 1)d$
$-8 = a + 2d$ ( \text{…(1)} )

For the 9th term:

$a_9 = a + (9 - 1)d$
$4 = a + 8d$ ( \text{…(2)} )

Subtract equation (1) from equation (2):

$(a + 8d) - (a + 2d) = 4 - (-8)$
$a + 8d - a - 2d = 4 + 8$
$6d = 12$
$d = \frac{12}{6}$
$d = 2$

The assertion states that $d = -2$ , which is incorrect. Therefore, the assertion (A) is false.

From our calculations, we have:

$(a + 8d) - (a + 2d) = 4 - (-8)$
$(a + 8d) - (a + 2d) = 12$

The reason claims:

$(a + 8d) - (a + 2d) = -8 - 4$
$(a + 8d) - (a + 2d) = -12$

This is incorrect as well. Thus, the reason (R) is false.

Hence, option 2 is the correct option.

Question 1(h)

The n^th term of a sequence = 5n^2 – 3.

Statement (1): The sequence is an A.P.

Statement (2): If the n^th term of a sequence is not linear, the sequence does not form an A.P.

(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (d) Statement 1 is false, and statement 2 is true.

We are given the expression for the n-th term of a sequence as $a_n = 5n^2 - 3$ .

Let’s calculate the first few terms:

\begin{align*}& a_1 = 5(1)^2 - 3 = 5 - 3 = 2, \\& a_2 = 5(2)^2 - 3 = 5 \times 4 - 3 = 20 - 3 = 17, \\& a_3 = 5(3)^2 - 3 = 5 \times 9 - 3 = 45 - 3 = 42.\end{align*}

Now, let’s find the differences between consecutive terms:

\begin{align*}& a_3 - a_2 = 42 - 17 = 25, \\& a_2 - a_1 = 17 - 2 = 15.\end{align*}

Notice that the differences between consecutive terms are not the same. ∴ this sequence is not an arithmetic progression (A.P.). This means statement 1 is false.

The general formula for an A.P. is given by ( a_n = a + (n – 1)d ), which is linear in terms of $n$ . Here, the given $a_n = 5n^2 - 3$ is quadratic, not linear.

∴ if $T_n$ is not a linear expression in $n$ , the sequence cannot be an A.P. Hence, statement 2 is true.

Therefore, option 4 is the correct option.

Question 1(i)

The sum of first 10 term of an A.P. = 3 and the sum of its first 15 term = 16.

Statement (1): The sum of first five terms of the given AP equals to 16 – 3 = 13.

Statement (2): The sum of last 5 terms of the given AP equals to sum of first 15 term minus sum of first 10 terms.

(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.

Answer: (d) Statement 1 is false, and statement 2 is true.

Consider an arithmetic progression (A.P.) where the first term is denoted by $a$ and the common difference by $d$ .

The formula for the sum of the first $n$ terms of an A.P. is given by:

S_n = \dfrac{n}{2}[2a + (n - 1)d]

We know the sum of the first 10 terms is 3:

S_{10} = 3

Substituting into the formula:

$\dfrac{10}{2}[2a + (10 - 1)d] = 3$
$\Rightarrow 5[2a + 9d] = 3$
$\Rightarrow 10a + 45d = 3 \quad \text{(Equation 1)}$

Also, the sum of the first 15 terms is 16:

$\dfrac{15}{2}[2a + (15 - 1)d] = 16$
$\Rightarrow \dfrac{15}{2}[2a + 14d] = 16$
$\Rightarrow 15[a + 7d] = 16$
$\Rightarrow 15a + 105d = 16 \quad \text{(Equation 2)}$

Subtracting Equation 1 from Equation 2:

$(15a + 105d) - (10a + 45d) = 16 - 3$
$\Rightarrow 15a + 105d - 10a - 45d = 13$
$\Rightarrow 5a + 60d = 13$

Now, let us determine the sum of the first 5 terms, $S_5$ :

$S_5 = \dfrac{5}{2}[2a + (5 - 1)d]$
$= \dfrac{5}{2}[2a + 4d]$
$= \dfrac{5}{2}[2(a + 2d)]$
$= 5[a + 2d]$
$= 5a + 10d$

Clearly, $5a + 10d$ is not equal to 13.

∴ Statement 1 is incorrect.

For the sum of the last 5 terms, consider terms from $a_{11}$ to $a_{15}$ :

$a_{11} + a_{12} + a_{13} + a_{14} + a_{15}$
$= (a + 10d) + (a + 11d) + (a + 12d) + (a + 13d) + (a + 14d)$
$= 5a + 60d$

We previously calculated:

\text{Sum of first 15 terms} - \text{Sum of first 10 terms} = 5a + 60d

Thus, the sum of the last 5 terms indeed equals the difference between the sum of the first 15 terms and the sum of the first 10 terms.

∴ Statement 2 is true.

Option 4 is the correct option.

Question 2

The 6^th term of an A.P. is 16 and the 14^th term is 32. Determine the 36^th term.

Answer:

Consider the first term of the arithmetic progression as $a$ and the common difference as $d$ .

We know:

a_6 = a + (6 - 1)d

This simplifies to:

a + 5d = 16 \quad \text{...(i)}

Similarly, for the 14th term:

a_{14} = a + (14 - 1)d

Which gives us:

a + 13d = 32 \quad \text{...(ii)}

Now, subtract equation (i) from equation (ii):

a + 13d - (a + 5d) = 32 - 16

This results in:

8d = 16

Solving for $d$ , we find:

d = 2

Substitute $d = 2$ back into equation (i):

a + 5(2) = 16

a + 10 = 16

So, we have:

a = 6

Now, to find the 36th term:

a_{36} = a + (36 - 1)d = 6 + (35)(2) = 76

Therefore, the 36th term is 76.

Question 3

If the third and the 9^th terms of an A.P. be 4 and -8 respectively, find which term is zero?

Answer:

Assume the first term of the arithmetic progression (A.P.) is $a$ and the common difference is $d$ .

Given:

For the third term, $a_3 = a + (3 - 1)d = a + 2d = 4$ ……..(i)

For the ninth term, $a_9 = a + (9 - 1)d = a + 8d = -8$ ……..(ii)

Now, subtract equation (i) from equation (ii):

a + 8d - (a + 2d) = -8 - 4

This simplifies to:

8d - 2d = -12

⇒ $6d = -12$

⇒ $d = -2$ .

Substitute $d = -2$ back into equation (i):

a + 2(-2) = 4

⇒ $a - 4 = 4$

⇒ $a = 8$ .

Now, find the term where the value is zero. Let the $n^{th}$ term be zero:

a_n = 0

⇒ $a + (n - 1)d = 0$

Substitute $a = 8$ and $d = -2$ :

8 + (n - 1)(-2) = 0

This simplifies to:

8 - 2n + 2 = 0

⇒ $2n = 10$

⇒ $n = 5$ .

Hence, the 5th term is zero.

Question 4

An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term of the A.P.

Answer:

Consider the first term of the arithmetic progression (A.P.) as $a$ and the common difference as $d$ .

We know:

a_3 = a + (3 - 1)d

This simplifies to:

a + 2d = 12 \quad \text{...(i)}

For the last term:

a_{50} = a + (50 - 1)d

Which gives:

a + 49d = 106 \quad \text{...(ii)}

To find $d$ , subtract equation (i) from equation (ii):

(a + 49d) - (a + 2d) = 106 - 12

Simplifying, we get:

49d - 2d = 94

47d = 94

d = 2

Now, substitute $d = 2$ back into equation (i):

a + 2(2) = 12

a + 4 = 12

a = 8

Next, to find the 29th term $a_{29}$ :

a_{29} = a + (29 - 1)d

a + 28d = 8 + 28(2) = 8 + 56 = 64

Thus, the 29th term is 64.

Question 5

Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.

Answer:

To find the sum of the first 20 terms of the arithmetic progression, we use the formula for the sum of an A.P.:

S = \dfrac{n}{2}(a + l)

where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term. Here, $n = 20$ , $a = 3$ , and $l = 57$ .

Substituting these values into the formula, we get:

S = \dfrac{20}{2}(3 + 57)

Simplifying further:

S = 10 \times 60

S = 600

Hence, sum of first 20 terms = 600.

Question 6

How many terms of the series 18 + 15 + 12 + ……. when added together will give 45?

Answer:

Consider the series where we need to find how many terms sum to 45. Here, the first term $a = 18$ and the common difference $d = 15 - 18 = -3$ .

The formula for the sum of $n$ terms of an arithmetic progression is:

S = \dfrac{n}{2}[2a + (n - 1)d]

Substituting the known values into the formula:

45 = \dfrac{n}{2}(2 \times 18 + (n - 1) \times (-3))

This simplifies to:

45 = \dfrac{n}{2}(36 - 3n + 3)

Further simplification gives:

45 = \dfrac{n}{2}(39 - 3n)

Multiplying through by 2 to eliminate the fraction:

90 = n(39 - 3n)

Reorganizing gives:

3n(13 - n) = 90

Dividing by 3, we have:

n(13 - n) = 30

Expanding and rearranging the quadratic equation:

n^2 - 13n + 30 = 0

To factorize, split the middle term:

n^2 - 10n - 3n + 30 = 0

Grouping terms, we get:

n(n - 10) - 3(n - 10) = 0

Factoring out the common factor:

(n - 3)(n - 10) = 0

Thus, the solutions are:

n = 3, 10

Hence, $n = 3$ or $n = 10$ .

Question 7

Find the general term (n^th term) and 23^rd term of the sequence, 3, 1, -1, -3, ………

Answer:

In this arithmetic progression, the common difference can be calculated as $1 - 3 = -2$ and the first term $a = 3$ .

The formula for the general term $a_n$ of an arithmetic progression is:

a_n = a + (n - 1) \, d

Substitute the known values:

a_n = 3 + (n - 1)(-2)

Simplify the expression:

= 3 - 2n + 2

= 5 - 2n

To find the 23rd term $a_{23}$ , substitute $n = 23$ into the general term:

a_{23} = 5 - 2(23)

= 5 - 46

= -41

Hence, $a_n = 5 - 2n$ and $a_{23} = -41$ .

Question 8

Is -150 a term of 11, 8, 5, 2, ………?

Answer:

Consider the sequence given: 11, 8, 5, 2, …. Notice that the difference between consecutive terms is constant:

8 – 11 = -3
5 – 8 = -3

Thus, this sequence forms an arithmetic progression (A.P.) with a common difference of -3.

Assume -150 is the n-th term of this A.P. We have:

∴ a_n = -150

Using the formula for the n-th term of an A.P., a + (n – 1)d = -150, where a is the first term and d is the common difference:

⇒ 11 + (-3)(n – 1) = -150

Simplifying this, we get:

⇒ 11 – 3n + 3 = -150

⇒ 14 – 3n = -150

⇒ -3n = -164

⇒ n = $\dfrac{164}{3} = 54\dfrac{2}{3}$ .

Since the number of terms in a sequence must be a whole number, -150 cannot be a term in this sequence.

Hence, -150 is not a term of 11, 8, 5, 2, ………

Question 9

How many multiples of 4 lie between 10 and 250?

Answer:

Consider the multiples of 4 that fall within the range of 10 to 250. These multiples are:

12, 16, 20, 24, ………., 248.

This sequence forms an arithmetic progression (A.P.) where the common difference $d = 4$ .

Assume 248 is the $n^{th}$ term of this A.P.

∴ $a_n = 248$

Using the formula for the $n^{th}$ term of an A.P., we have:

12 + 4(n - 1) = 248

Simplifying this equation:

12 + 4n - 4 = 248

\Rightarrow 4n + 8 = 248

\Rightarrow 4n = 240

$\Rightarrow n = 60$ .

Therefore, there are 60 multiples of 4 between 10 and 250.

Question 10

The 25^th term of an A.P. exceeds its 9^th term by 16. Find its common difference.

Answer:

We know that the 25 $^{th}$ term of an arithmetic progression exceeds the 9 $^{th}$ term by 16. This can be expressed as:

⇒ $a_{25} - a_{9} = 16$

Substitute the formula for the general term of an A.P., (a_n = a + (n-1)d), into the equation:

⇒ (a + (25 – 1)d – [a + (9 – 1)d] = 16)

Simplifying the expression, we have:

⇒ $a - a + 24d - 8d = 16$

⇒ $16d = 16$

Dividing both sides by 16 gives:

⇒ $d = 1$ .

Hence, common difference = 1.

Question 11

If the n^th term of the A.P. 58, 60, 62, …… is equal to the n^th term of the A.P. -2, 5, 12, ……., find the value of n.

Answer:

To find the n^th term of the arithmetic progression (A.P.) 58, 60, 62, …, let’s first identify the common difference. Notice that:

$60 - 58 = 2$
$62 - 60 = 2$

Thus, the sequence is an A.P. with a common difference of 2. The formula for the n^th term is:

a_n = a + (n - 1)d

Substituting the values, we have:

$a_n = 58 + (n - 1) \times 2$
$= 58 + 2n - 2$
$= 2n + 56$

Now, consider the n^th term of the A.P. -2, 5, 12, …. Here, calculate the common difference:

$5 - (-2) = 7$
$12 - 5 = 7$

This sequence is also an A.P. with a common difference of 7. The n^th term formula is:

a_n = a + (n - 1)d

Plugging in the values, we find:

$a_n = -2 + (n - 1) \times 7$
$= -2 + 7n - 7$
$= 7n - 9$

Since the n^th terms of both sequences are equal, set the expressions equal:

2n + 56 = 7n - 9

Solving for n:

$7n - 2n = 56 + 9$
$5n = 65$
$n = 13$

Hence, n = 13.

Question 12

Which term of the A.P. 105, 101, 97, ………, is the first negative term?

Answer:

Consider the term in the sequence where the value becomes negative. Denote this term as $a_n$ .

∴ $a_n < 0$

The general formula for the $n^{th}$ term of an arithmetic progression is given by:

a + (n - 1)d < 0

Substitute the values from the sequence:

105 + (n - 1)(-4) < 0

Simplifying, we get:

105 - 4n + 4 < 0

Combine like terms:

109 - 4n < 0

Rearrange to isolate $n$ :

4n > 109

Divide both sides by 4:

n > \dfrac{109}{4} = 27\dfrac{1}{4}

Since $n$ must be an integer, the smallest integer greater than $27\dfrac{1}{4}$ is 28.

Thus, the 28th term is the first negative term.

Question 13

Divide 216 into three parts which are in A.P. and the product of two smaller parts is 5040.

Answer:

Consider three numbers in an arithmetic progression: $a - d$ , $a$ , and $a + d$ . We are given that their sum is 216.

∴ $(a - d) + a + (a + d) = 216$

Simplifying, we have:

⇒ $3a = 216$

⇒ $a = 72$ .

Next, we know the product of the two smaller numbers is 5040:

⇒ $(a - d) \times a = 5040$

Substituting $a = 72$ :

⇒ $(72 - d) \times 72 = 5040$

⇒ $5184 - 72d = 5040$

Solving for $d$ :

⇒ $72d = 5184 - 5040$

⇒ $72d = 144$

⇒ $d = 2$ .

Thus, the numbers are $72 - 2$ , $72$ , and $72 + 2$ .

Hence, numbers are 70, 72, 74.

Question 14

Can 2n^2 – 7 be the n^th term of an A.P. Explain.

Answer:

Consider the given expression for the n-th term:

a_n = 2n^2 - 7

Let’s calculate the first few terms:

For $n = 1$ , ( a_1 = 2(1)^2 – 7 = 2 – 7 = -5 )
For $n = 2$ , ( a_2 = 2(2)^2 – 7 = 8 – 7 = 1 )
For $n = 3$ , ( a_3 = 2(3)^2 – 7 = 18 – 7 = 11 )

Now, check the differences between consecutive terms:

( a_2 – a_1 = 1 – (-5) = 6 )
$a_3 - a_2 = 11 - 1 = 10$

Notice that these differences are not the same. ∴ the sequence does not have a common difference.

Hence, 2n^2 – 7 cannot be the n^th term of an A.P.

Question 15

The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^st term of this A.P.

Answer:

To find the sum of the first seven terms in an arithmetic progression, we use the formula:

S = \dfrac{n}{2}[2a + (n - 1)d]

Given that the sum $S$ is $2100$ , and substituting the known values:

$2100 = \dfrac{7}{2}$ 2 \times 20 + (7 – 1)d$$$$

Simplifying inside the brackets, we have:

2100 = \dfrac{7}{2}[40 + 6d]

By factoring a $2$ out of the terms in the bracket, we can cancel the fraction:

2100 = 7(20 + 3d)

Now, divide by $7$ :

300 = 20 + 3d

Subtract $20$ from both sides:

3d = 280

Finally, divide by $3$ to find $d$ :

d = \dfrac{280}{3}

Next, to find the $31$ st term $a_{31}$ , use the formula:

a_n = a + (n - 1)d

So, the 31st term is:

a_{31} = 20 + (31 - 1) \times \dfrac{280}{3}

Calculating further:

= 20 + 30 \times \dfrac{280}{3}

= 20 + 2800

= 2820

Hence, the 31st term of the A.P. is 2820.

Question 16

Find the sum of last 8 terms of the A.P.

-12, -10, -8, …….., 58.

Answer:

To find the sum of the last 8 terms of the arithmetic progression given by -12, -10, -8, …, 58, we can reverse the sequence to consider the first 8 terms of a new A.P.: 58, 56, 54, …, -10, -12.

For this A.P., the sum of the first 8 terms is calculated using the formula:

S = \dfrac{n}{2}[2a + (n - 1)d]

Substitute the values:

Number of terms $n = 8$
First term $a = 58$
Common difference $d = -2$

Thus, we have:

S = \dfrac{8}{2}[2 \times 58 + (8 - 1) \times (-2)]

Simplifying further:

= 4[116 - 14]

= 4 \times 102

= 408.

Therefore, the sum is 408.

Question 17

An A.P. consists of 57 terms of which 7^th term is 13 and the last term is 138. Find the 45^th term of this A.P.

Answer:

The arithmetic progression has a total of 57 terms.

We know the formula for the $n^{th}$ term of an A.P. is:

a_n = a + (n - 1) d

For the $7^{th}$ term:

a_7 = a + 6d

Given that $a_7 = 13$ , we have:

a + 6d = 13 \quad \text{...(1)}

For the last term, which is the $57^{th}$ term:

a_{57} = a + 56d

Given that $a_{57} = 138$ , we get:

a + 56d = 138 \quad \text{...(2)}

Now, subtract equation (1) from equation (2):

(a + 56d) - (a + 6d) = 138 - 13

50d = 125

Solving for $d$ gives:

d = \frac{125}{50}

d = 2.5

Substitute $d = 2.5$ back into equation (1):

a + 6(2.5) = 13

a + 15 = 13

a = 13 - 15

a = -2

To find the $45^{th}$ term:

a_{45} = a + (45 - 1)d

a_{45} = -2 + (45 - 1)(2.5)

a_{45} = -2 + 110

a_{45} = 108

Therefore, the $45^{th}$ term of this A.P. is 108.

Question 18

Ten times the tenth term of an A.P. is equal to fifteen times its fifteenth term. Find the twenty-fifth term of this A.P.

Answer:

Consider the first term of an arithmetic progression (A.P.) as $a$ and the common difference as $d$ .

The formula for the $n$ -th term of an A.P. is given by:

a_n = a + (n - 1)d

According to the problem, ten times the tenth term is equal to fifteen times the fifteenth term. This gives us:

10a_{10} = 15a_{15}

Substituting the formula for $a_n$ :

10(a + 9d) = 15(a + 14d)

Expanding both sides, we have:

10a + 90d = 15a + 210d

Rearranging the terms, we get:

10a - 15a + 90d - 210d = 0

Simplifying further:

-5a - 120d = 0

This can be rewritten as:

5a + 120d = 0

Factoring out the common factor:

5(a + 24d) = 0

Thus, we have:

a + 24d = 0

Solving for $a$ , we find:

$a = -24d$ (\text{…..(1)})

Now, to find the 25th term:

a_{25} = a + (25 - 1)d

Substitute the value of $a$ from equation (1):

a_{25} = -24d + 24d

This simplifies to:

a_{25} = 0

Hence, the twenty-fifth term of this A.P. = 0

Question 19

The sum of the 2^nd term and the 7^th term of an A.P. is 30. If its 15^th term is 1 less than twice of its 8^th term, find this A.P.

Answer:

Consider the first term as $a$ and the common difference as $d$ .

From the problem statement, we have:

∴ $a_2 + a_7 = 30$

This implies:

⇒ $a + (2 - 1)d + a + (7 - 1)d = 30$

⇒ $a + d + a + 6d = 30$

⇒ $2a + 7d = 30$ ………(i)

Additionally, we know:

∴ $2a_8 - 1 = a_{15}$

This translates to:

⇒ $2[a + (8 - 1)d] - 1 = a + (15 - 1)d$

⇒ $2[a + 7d] - 1 = a + 14d$

⇒ $2a + 14d - 1 = a + 14d$

Solving this, we get:

⇒ $2a - a - 1 = 14d - 14d$

⇒ $a - 1 = 0$

⇒ $a = 1$ .

Now, substituting $a = 1$ in equation (i):

⇒ $2(1) + 7d = 30$

⇒ $2 + 7d = 30$

⇒ $7d = 28$

⇒ $d = 4$ .

Thus, the A.P. is given by $a, (a + d), (a + 2d),\ldots$

= $1, (1 + 4), (1 + 2.4),\ldots$

= $1, 5, 9,\ldots$

Hence, A.P. = 1, 5, 9,\ldots

Question 20

Refer the given sequence 23, $21\dfrac{1}{2}$ , 20, ….

(a) Find the general term of the given sequence.

(b) Which term is the last positive term in the sequence.

Answer:

(a) Consider the arithmetic progression: 23, $21\dfrac{1}{2}$ , 20, …

Here, the first term $a = 23$ and the common difference $d$ is calculated as follows:

d = 21\dfrac{1}{2} - 23 = -1\dfrac{1}{2} = -\dfrac{3}{2}.

The formula for the general term $a_n$ of an arithmetic progression is given by:

a_n = a + (n - 1)d.

Substituting the values we have:

\begin{aligned}a_n = 23 - \left(\dfrac{3}{2}\right)(n - 1) \\\Rightarrow a_n = 23 - \left(\dfrac{3n - 3}{2}\right) \\\Rightarrow a_n = \left(\dfrac{46 - 3n + 3}{2}\right) \\\Rightarrow a_n = \left(\dfrac{49 - 3n}{2}\right).\end{aligned}

Thus, the general term $a_n$ is $\left(\dfrac{49 - 3n}{2}\right)$ .

(b) To find the last positive term, we need $a_n > 0$ :

\dfrac{49 - 3n}{2} > 0.

Solving this inequality:

\begin{aligned}49 - 3n > 0 \\\Rightarrow 3n < 49 \\\Rightarrow n < \dfrac{49}{3} \\\Rightarrow n < 16\dfrac{1}{3}.\end{aligned}

This means the largest integer value for $n$ is 16. Therefore, the 16th term is the last positive term.

Calculating $a_{16}$ :

a_{16} = \left(\dfrac{49 - 3 \times 16}{2}\right) = \dfrac{49 - 48}{2} = \dfrac{1}{2} = 0.5.

Hence, the 16th term is the last positive term.

Case-Study Based Questions

Question 1

Case study:
Cable cars, at hill stations, are major tourist attractions. On a hill station, the length of a cable car ride from the base to the topmost point on the hill is 5000 m. Poles are installed at equal intervals on the way to provide support to the cable on which the car moves.

The distance of the first pole from the base point is 200 m and subsequent poles are installed at equal intervals of 150 m. Further, the distance of the last pole from the top is 300 m.

Based on above information, answer the following questions using Arithmetic Progression:

(i) Find the distance of the 10^th pole from the base.

(ii) Find the distance between the 15^th pole and 25^th pole.

(iii) Find the time taken by the cable car to reach the 15^th pole from the top if it is moving at a speed of 5 m/s and coming from the top.

Answer:

(i) We’re given the total cable length of 5000 m. The first pole is positioned 200 m from the base, and poles are spaced 150 m apart. Thus, the sequence of distances forms an arithmetic progression: 200, 350, 500, 650, and so on.

To find the distance of the 10th pole from the base, apply the formula for the nth term of an A.P.:

a_n = a + (n - 1)d

Here, $a = 200$ , $d = 150$ , and $n = 10$ .

⇒ $a_{10} = 200 + (9) \times 150$

= $200 + 1350$

= $1550$ m.

Thus, the 10th pole is 1550 m from the base.

(ii) To determine the distance between the 15th and 25th poles, we use the same formula.

For the 15th pole:

a_{15} = 200 + (14) \times 150

= $200 + 2100$

= $2300$ m.

For the 25th pole:

a_{25} = 200 + (24) \times 150

= $200 + 3600$

= $3800$ m.

The distance between the 15th and 25th poles is:

$3800 - 2300 = 1500$ m.

Thus, the distance between these poles is 1500 m.

(iii) To find the time taken to reach the 15th pole from the top, we first need to determine the total number of poles.

Since the last pole is 300 m from the top of the 5000 m cable, its distance from the base is $5000 - 300 = 4700$ m.

Using the formula $a_n = a + (n - 1)d$ , set $a_n = 4700$ .

⇒ $200 + (n - 1) \times 150 = 4700$

⇒ $150(n - 1) = 4500$

⇒ $n - 1 = \dfrac{4500}{150}$

⇒ $n - 1 = 30$

⇒ $n = 31$ .

Thus, there are 31 poles in total.

The 15th pole from the top is the 17th pole from the base, calculated as $(31 - 15 + 1) = 17$ .

The distance from the 31st to the 17th pole is $(31 - 17) \times 150 = 14 \times 150 = 2100$ m.

Adding the 300 m from the top to the last pole gives a total distance of $2100 + 300 = 2400$ m.

At a speed of 5 m/s, the time taken is:

$\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{2400}{5}$ = 480 seconds or 8 minutes.

Therefore, the time to reach the 15th pole from the top is 8 minutes.

Question 2

Case study:
A school auditorium is to be constructed to accommodate at least 1500 people. The chairs are to be placed in a concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

(i) If the first circular row has 30 seats, how many seats will the 10^th row have?

(ii) For 1500 seats in the auditorium, how many circular rows need to be there?

(iii) If there were 17 rows in the auditorium, how many seats will there be in the middle row?

Answer:

(i) We start with the first circular row having 30 seats, denoted by $a = 30$ . Each subsequent row increases by 10 seats, which means the common difference $d = 10$ . This sequence forms an arithmetic progression (A.P.): 30, 40, 50, 60, and so forth.

To find the number of seats in the 10th row, use the nth term formula of an A.P.:

a_n = a + (n - 1)d

Substitute the values:

$a_{10} = 30 + (10 - 1) \times 10$
$= 30 + 90$
$= 120$

Thus, the 10th row contains 120 seats.

(ii) To determine how many rows are needed for a total of 1500 seats, let $n$ represent the number of rows. The sum of an A.P. is given by:

S_n = \dfrac{n}{2}[2a + (n - 1)d]

Substitute the known values:

$1500 = \dfrac{n}{2}[2(30) + 10n - 10]$
$\Rightarrow 1500 = \dfrac{n}{2}[60 + 10n - 10]$
$\Rightarrow 1500 = \dfrac{n}{2}[50 + 10n]$
$\Rightarrow 3000 = 10n[5 + n]$
$\Rightarrow \dfrac{3000}{10} = n(n + 5)$
$\Rightarrow 300 = n(n + 5)$
$\Rightarrow 300 = n^2 + 5n$
$\Rightarrow n^2 + 5n - 300 = 0$

Solving this quadratic equation:

$n^2 + 20n - 15n - 300 = 0$
$\Rightarrow n(n + 20) - 15(n + 20) = 0$
$\Rightarrow (n - 15)(n + 20) = 0$
$\Rightarrow n - 15 = 0 \text{ or } n + 20 = 0$
$\Rightarrow n = -20 \text{ or } n = 15$

Since the number of rows cannot be negative, we have $n = 15$ .

Therefore, 15 rows are needed to accommodate 1500 seats.

(iii) With 17 rows in total, the middle row is found by calculating:

Since 17 is an odd number:

\text{Middle row number} = \dfrac{17 + 1}{2} = \dfrac{18}{2} = 9

To find the number of seats in the 9th row, use the nth term formula again:

$a_9 = a + (n - 1)d$
$\Rightarrow a_9 = 30 + (9 - 1) \times 10$
$= 30 + 80$
$= 110$

Therefore, the middle row contains 110 seats.

Question 3

Case study:
The figure shows a big triangle in which multiple other triangles can be seen. Observe the pattern of dark shaded and light unshaded triangles starting with one triangle in row 1, three triangles in row 2, five triangles in row 3 and so on.

Based on the above information, answer the following questions:

(i) How many triangles will be there in the 15^th row?

(ii) In which row will the number of triangles be 47?

(iii) The number of dark shaded triangles in each row are in A.P. Find the total number of dark shaded triangles in the first 15 rows.

Answer:

(i) The sequence of triangles in each row forms an arithmetic progression (A.P.):

1, 3, 5, 7, 9, …

Here, the first term $a = 1$ and the common difference $d = 2$ . To find the number of triangles in the 15th row, use the formula for the $n^{th}$ term of an A.P.:

a_n = a + (n - 1) \cdot d

Substituting the values, we have:

$a_{15} = 1 + (15 - 1) \times 2$
$= 1 + 14.2$
$= 1 + 28$
$= 29$

∴ There are 29 triangles in the 15th row.

(ii) To determine the row number where there are 47 triangles, let this row be the $n^{th}$ row. Using the formula:

47 = 1 + (n - 1) \cdot 2

Rearranging gives:

$47 - 1 = 2(n - 1)$
$\frac{46}{2} = n - 1$
$23 = n - 1$
$n = 24$

∴ The 47 triangles are in the 24th row.

(iii) The number of dark shaded triangles in each row also follows an A.P. with $a = 1$ and $d = 1$ . We need to find the sum of the first 15 terms:

The sum of an A.P. is given by:

$S_n = \frac{n}{2}$ 2a + (n – 1) \cdot d$$$$

Substituting the values, we calculate:

$S_{15} = \frac{15}{2}$ 2(1) + (15 – 1) \times 1 $$$$ = \frac{15}{2} \times [2 + 14]= \frac{15}{2} \times 16= 15 \times 8= 120$$

∴ The total number of dark shaded triangles in the first 15 rows is 120.

Frequently Asked Questions

The formula to find the nth term (an) of an Arithmetic Progression is an = a + (n-1)d, where 'a' is the first term, 'n' is the term number, and 'd' is the common difference of the sequence. This formula is essential for finding any specific term in the series without listing all the preceding terms.

There are two main formulas to find the sum of the first 'n' terms (Sn). The first is Sn = n/2 * [2a + (n-1)d], where 'a' is the first term and 'd' is the common difference. The second formula, used when the last term (l) is known, is Sn = n/2 * (a + l).

The common difference, denoted by 'd', is the constant value that is added to each term to get the next term in an Arithmetic Progression. It can be found by subtracting any term from its succeeding term (e.g., d = a2 – a1). The common difference can be positive, negative, or zero.

Yes, all solutions for the Arithmetic Progression chapter are prepared by subject matter experts following the exact steps and format required by the ICSE board. This ensures that you not only get the correct answer but also learn the proper method to present your solutions in the board exams for full marks.

ICSE Board

Exercise 10(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10(i)

Question 10(ii)

Question 10(iii)

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Exercise 10(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Exercise 10(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4(i)

Question 4(ii)

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11(i)

Question 11(ii)

Question 12

Question 13

Exercise 10(D)

Question 1(a)

Question 1(b)

Question 1(c)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7