Concise Mathematics Selina Solutions Class 10 ICSE Chapter 11 Geometric Progression

Our detailed Selina ICSE Class 10 Geometric Progression Solutions provide a clear guide to understanding sequences where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This chapter from your Class 10 Concise Mathematics Selina textbook builds upon your knowledge of Arithmetic Progressions, introducing a new type of sequence with fascinating properties. You will learn the fundamental formulas to find the n^th term of a G.P., calculate the sum of a finite number of terms, and even explore the concept of the sum of an infinite geometric series. Mastering these concepts is crucial as they form the basis for many applications in finance, science, and higher-level mathematics.

If you are looking for clear, step-by-step solutions for any of the 73 questions in the Geometric Progression chapter, you have come to the right place. We understand that you might be stuck on a specific problem or want to verify that your method is correct before an exam. Each solution provided here for Exercise 11(A), Exercise 11(B), Test Yourself, and the Case Study question follows the exact methodology prescribed by the ICSE board. Here you will find reliable, expertly solved problems to help you master every concept and secure better marks in your Maths exam.

Exercise 11(A)

Question 1(a)

If the first term of a G.P. is 8 and its common ratio is -2. The 3^rd term of this G.P. is :

• (a) -32
• (b) 2
• (c) 32
• (d) -2

We have a geometric progression where the initial term is 8 and the common ratio is -2.

The formula to find the nth term of a G.P. is:

a_n = ar^{n-1}

For the 3rd term, substitute the given values:

a_3 = 8 \times (-2)^{3-1}

This simplifies to:

a_3 = 8 \times (-2)^2

Calculating further, we have:

a_3 = 8 \times 4 a_3 = 32

Hence, Option 3 is the correct option.

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Question 1(b)

The 4^th term of a G.P. is 16 and the 7^th term is 128, then its common ratio is equal to :

• (a) 2
• (b) -2
• (c) 1
• (d) -1

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

The formula for the n^{th} term of a G.P. is given by:

⇒ a_n = ar^{n-1}

We know the 4^{th} term is 16:

⇒ a_4 = 16

⇒ ar^{4-1} = 16

⇒ ar^3 = 16 ………(1)

Also, the 7^{th} term is 128:

⇒ a_7 = 128

⇒ ar^{7-1} = 128

⇒ ar^6 = 128 ………(2)

Now, divide equation (2) by equation (1):

\begin{aligned}\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{128}{16} \\\Rightarrow r^3 = 8 \\\Rightarrow r^3 = 2^3 \\\Rightarrow r = 2.\end{aligned}

Hence, Option 1 is the correct option.

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Question 1(c)

(2x + 2) and (3x + 3) are two consecutive terms of a G.P. The common ratio of the G.P. is :

• (a) 2
• (b) 3
• (c) \dfrac{3}{2}
• (d) \dfrac{2}{3}

To find the common ratio in a geometric progression, we use the formula:

Common ratio = \dfrac{a_{n + 1}}{a_n}

Here, the terms given are (2x + 2) and (3x + 3), which are consecutive terms of the G.P.

∴ Common ratio = \dfrac{3x + 3}{2x + 2}

⇒ Simplifying this, we get:

= \dfrac{3(x + 1)}{2(x + 1)}

⇒ Canceling (x + 1) from numerator and denominator, we have:

= \dfrac{3}{2}.

Hence, Option 3 is the correct option.

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Question 1(d)

The third term of a G.P. is 3. The product of its first five terms is :

• (a) 15
• (b) \dfrac{3 \times 5}{2}
• (c) 3^5
• (d) 5^3

Consider the sequence of the first five terms of a geometric progression (G.P.) as follows:

\dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2

We know that the third term of this G.P. is given as 3. Therefore, we have:

∴ a = 3

Now, let’s find the product of these five terms:

\Rightarrow \dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2

When we simplify the expression, all the r terms cancel out, leaving us with:

\Rightarrow a^5

Since a = 3, it follows that:

\Rightarrow 3^5

Hence, Option 3 is the correct option.

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Question 1(e)

The 8^th term of a G.P. is 192 and its common ratio is 2, then the first term of the G.P. is :

• (a) 3
• (b) \dfrac{1}{3}
• (c) \dfrac{2}{3}
• (d) \dfrac{3}{2}

Suppose the first term of the geometric progression is denoted by a and the common ratio by r. Using the formula for the n^{th} term of a G.P., we have:

a_n = ar^{n-1}

Given that the 8^{th} term of the G.P. is 192, we can write:

a_8 = 192

This implies:

ar^{8-1} = 192 ar^7 = 192

Given the common ratio r = 2, substitute this value into the equation:

a(2)^7 = 192 128a = 192

Solving for a gives:

a = \frac{192}{128} = \frac{3}{2}

Hence, option 4 is the correct option.

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Question 2

Find the 9^th term of the series :

1, 4, 16, 64, ……..

Observe that the ratio of successive terms is consistent: \frac{4}{1} = \frac{16}{4} = 4. This indicates the series is a geometric progression (G.P.) with the first term a = 1 and common ratio r = 4.

For a G.P., the formula for the n^\text{th} term is given by:

a_n = ar^{n-1}

Substituting the values for the 9th term, we have:

a_9 = 1 \times (4)^{9-1} = (4)^8 = 65536

Thus, the 9th term of the series is 65536.

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Question 3

Find the seventh term of the G.P. :

1, \sqrt{3}, 3, 3\sqrt{3}, ........

Observe that the ratio between consecutive terms is constant:

\frac{\sqrt{3}}{1} = \frac{3\sqrt{3}}{3} = \sqrt{3}

Thus, this series is a geometric progression (G.P.) where the common ratio r is \sqrt{3} and the first term a is 1.

Recall the formula for the n^{th} term of a G.P., which is given by:

a_n = ar^{n-1}

For the 7th term, ( a_7 = 1 \times (\sqrt{3})^{7-1} ).

This simplifies to:

(\sqrt{3})^6

Calculating further, we find:

(\sqrt{3})^6 = 27

Therefore, the 7th term of the G.P. is 27.

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Question 4

Find the 8^th term of the sequence :

\dfrac{3}{4}, 1\dfrac{1}{2}, 3, .........

The sequence provided is \dfrac{3}{4}, \dfrac{3}{2}, 3, \ldots

To determine if this is a geometric progression, let’s find the common ratio between consecutive terms.

Calculate the ratio of the second term to the first term:

\dfrac{\dfrac{3}{2}}{\dfrac{3}{4}} = \dfrac{3 \times 4}{2 \times 3} = 2.

Similarly, for the third term to the second term:

\dfrac{3}{\dfrac{3}{2}} = 2.

Both ratios are equal to 2, confirming that the sequence is a geometric progression (G.P.) with the first term a = \dfrac{3}{4} and common ratio r = 2.

The formula for the n^{th} term of a G.P. is:

a_n = ar^{n-1}

To find the 8th term (a_8), substitute the known values:

a_8 = \dfrac{3}{4} \times (2)^{8-1} = \dfrac{3}{4} \times 2^7 = \dfrac{3}{4} \times 128 = 3 \times 32

= 96.

Thus, the 8th term a_8 = 96.

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Question 5

Find the next three terms of the sequence :

\sqrt{5}, 5, 5\sqrt{5}, .......

Observe that:

\dfrac{5}{\sqrt{5}} = \dfrac{5\sqrt{5}}{5} = \sqrt{5}

This confirms that the sequence \sqrt{5}, 5, 5\sqrt{5}, \ldots is a geometric progression (G.P.) with the common ratio r = \sqrt{5} and the first term a = \sqrt{5}.

The next three terms we need are the 4th, 5th, and 6th terms.

Recall the formula for the n^{th} term of a G.P.:

a_n = ar^{n-1}

Thus, for the 4th term:

a_4 = ar^{4-1} = ar^3 = \sqrt{5}(\sqrt{5})^3 = \sqrt{5}(5\sqrt{5}) = 25

For the 5th term:

a_5 = ar^{5-1} = ar^4 = \sqrt{5}(\sqrt{5})^4 = \sqrt{5}(25) = 25\sqrt{5}

And for the 6th term:

a_6 = ar^{6-1} = ar^5 = \sqrt{5}(\sqrt{5})^5 = \sqrt{5}(25\sqrt{5}) = 125

Therefore, the next three terms of the G.P. are 25, 25\sqrt{5}, and 125.

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Question 6

Find the seventh term of the G.P. :

\sqrt{3} + 1, 1, \dfrac{\sqrt{3} - 1}{2}, ………..

To simplify the term \dfrac{\sqrt{3} - 1}{2}, let’s rationalize it:

\Rightarrow \dfrac{\sqrt{3} - 1}{2} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1}

This gives us:

= \dfrac{\sqrt{3}^2 - 1^2}{2(\sqrt{3} + 1)} = \dfrac{3 - 1}{2(\sqrt{3} + 1)} = \dfrac{2}{2(\sqrt{3} + 1)} = \dfrac{1}{\sqrt{3} + 1}.

Thus, the sequence becomes \sqrt{3} + 1, 1, \dfrac{1}{\sqrt{3} + 1}, \ldots

The common ratio r is \dfrac{1}{\sqrt{3} + 1}.

According to the formula for the n^{th} term of a G.P.,

a_n = ar^{n-1}

For the seventh term, we have:

a_7 = (\sqrt{3} + 1)\left(\dfrac{1}{\sqrt{3} + 1}\right)^{7 - 1} = (\sqrt{3} + 1)\left(\dfrac{1}{\sqrt{3} + 1}\right)^{6} = \left(\dfrac{1}{\sqrt{3} + 1}\right)^{5}

Next, rationalize \dfrac{1}{\sqrt{3} + 1}:

= \left(\dfrac{1}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\right)^{5} = \left(\dfrac{\sqrt{3} - 1}{\sqrt{3}^2 - (1)^2}\right)^5 = \left(\dfrac{\sqrt{3} - 1}{3 - 1}\right)^5 = \left(\dfrac{\sqrt{3} - 1}{2}\right)^5 = \dfrac{1}{32}(\sqrt{3} - 1)^5.

Hence, seventh term of the G.P. = \dfrac{1}{32}(\sqrt{3} - 1)^5.

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Question 7

Find the next two terms of the series :

2 – 6 + 18 – 54 ………….

Observe that the ratio between consecutive terms is consistent:

\frac{-6}{2} = \frac{18}{-6} = -3.

Thus, this sequence is a Geometric Progression (G.P.) with the common ratio r = -3 and the first term a = 2.

Recall the formula for the n^{th} term of a G.P.:

a_n = ar^{n-1}

We need to find the 5th and 6th terms.

For the 5th term:

a_5 = ar^4 = 2 \times (-3)^4 = 2 \times 81 = 162.

For the 6th term:

a_6 = ar^5 = 2 \times (-3)^5 = 2 \times (-243) = -486.

Hence, the next two terms are 162 and -486.

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Question 8

Which term of the G.P. :

-10, \dfrac{5}{\sqrt{3}}, -\dfrac{5}{6}, ........ \text{ is } -\dfrac{5}{72}?

To find the common ratio r of the given geometric progression, we divide the second term by the first term:

r = \dfrac{\dfrac{5}{\sqrt{3}}}{-10} = -\dfrac{5}{10\sqrt{3}} = -\dfrac{1}{2\sqrt{3}}.

We need to determine which term is -\dfrac{5}{72}. Let’s denote the n^{th} term as -\dfrac{5}{72}.

The formula for the n^{th} term of a G.P. is given by ar^{n-1} = -\dfrac{5}{72}, where a is the first term. Substituting the values, we have:

-10 \times \left(-\dfrac{1}{2\sqrt{3}}\right)^{n-1} = -\dfrac{5}{72}.

Simplifying, we get:

\left(-\dfrac{1}{2\sqrt{3}}\right)^{n-1} = -\dfrac{5}{72} \times -\dfrac{1}{10}.

This simplifies to:

\left(-\dfrac{1}{2\sqrt{3}}\right)^{n-1} = \dfrac{1}{144}.

Recognizing the right side as a power of our common ratio:

\left(-\dfrac{1}{2\sqrt{3}}\right)^{n-1} = \left(-\dfrac{1}{2\sqrt{3}}\right)^4.

Thus, it follows that n - 1 = 4, leading to n = 5.

Therefore, the 5^{th} term of the G.P. is -\dfrac{5}{72}.

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Question 9

The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

From the problem, we know:

∴ a_5 = 81

⇒ ar^4 = 81 ……..(i)

And also:

∴ a_2 = 24

⇒ ar = 24 ……..(ii)

By dividing equation (i) by equation (ii), we have:

\begin{aligned}\Rightarrow \dfrac{ar^4}{ar} = \dfrac{81}{24} \\\Rightarrow r^3 = \dfrac{27}{8} \\\Rightarrow r^3 = \Big(\dfrac{3}{2}\Big)^3 \\\Rightarrow r = \dfrac{3}{2}.\end{aligned}

Now, substituting the value of r in equation (ii):

\begin{aligned}\Rightarrow a \times \dfrac{3}{2} = 24 \\\Rightarrow a = \dfrac{2}{3} \times 24 \\\Rightarrow a = 16.\end{aligned}

Next, calculate the third term:

∴ a_3 = ar^2

= 16 \times \Big(\dfrac{3}{2}\Big)^2

= 16 \times \dfrac{9}{4}

= 4 × 9

= 36.

Now, find the fourth term:

∴ a_4 = ar^3

= 16 \times \Big(\dfrac{3}{2}\Big)^3

= 16 \times \dfrac{27}{8}

= 2 × 27

= 54.

Thus, the G.P. is 16, 24, 36, 54, 81, ………..

Hence, G.P. = 16, 24, 36, 54, 81, ………..

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Question 10

Fourth and seventh terms of a G.P. are \dfrac{1}{18} \text{ and } -\dfrac{1}{486} respectively. Find the G.P.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

We know:

a_4 = \dfrac{1}{18}

This gives us:

ar^3 = \dfrac{1}{18} \quad \text{...(i)}

Additionally:

a_7 = -\dfrac{1}{486}

This implies:

ar^6 = -\dfrac{1}{486} \quad \text{...(ii)}

By dividing equation (ii) by equation (i), we have:

\dfrac{ar^6}{ar^3} = \dfrac{-\dfrac{1}{486}}{\dfrac{1}{18}}

Simplifying gives:

r^3 = -\dfrac{18}{486} r^3 = -\dfrac{1}{27} r^3 = \left(-\dfrac{1}{3}\right)^3

Thus, the common ratio is:

r = -\dfrac{1}{3}

Substituting r = -\dfrac{1}{3} back into equation (i):

a \times \left(-\dfrac{1}{3}\right)^3 = \dfrac{1}{18} a \times -\dfrac{1}{27} = \dfrac{1}{18}

Solving for a:

a = -\dfrac{27}{18} = -\dfrac{3}{2}

Now, let’s find the subsequent terms:

a_2 = ar = -\dfrac{3}{2} \times \left(-\dfrac{1}{3}\right) = \dfrac{1}{2} a_3 = ar^2 = -\dfrac{3}{2} \times \left(-\dfrac{1}{3}\right)^2 = -\dfrac{3}{2} \times \dfrac{1}{9} = -\dfrac{1}{6}

Therefore, the G.P. is:

-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, \dfrac{1}{18} \ldots

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Question 11

If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.

Consider the first term of the G.P. as a and the common ratio as r.

Given:

∴ a = 2

The third term of the G.P. is a r^2 = 8.

∴ 2r^2 = 8

⇒ r^2 = 4

⇒ r = \sqrt{4} = ±2

The second term, a_2, is given by ar.

If r = -2:

⇒ a_2 = 2(-2) = -4

If r = 2:

⇒ a_2 = 2(2) = 4

Hence, a_2 = 4 or -4.

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Question 12

The product of 3^rd term and 8^th terms of a G.P. is 243. If its 4^th term is 3, find its 7^th term.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

Given conditions are:

∴ The product of the 3rd and 8th terms is 243:

⇒ a_3 \times a_8 = 243

⇒ ar^2 \times ar^7 = 243

⇒ a^2r^9 = 243 ……….(i)

Additionally, the 4th term is given as 3:

⇒ a_4 = 3

⇒ ar^3 = 3

⇒ a = \dfrac{3}{r^3} ……..(ii)

Substitute the expression for a from equation (ii) into equation (i):

\Rightarrow \Big(\dfrac{3}{r^3}\Big)^2 \times r^9 = 243 \Rightarrow \dfrac{9}{r^6} \times r^9 = 243 \Rightarrow 9r^3 = 243 \Rightarrow r^3 = 27 \Rightarrow r = \sqrt[3]{27} \Rightarrow r = 3.

Now, substitute r = 3 back into equation (ii):

\Rightarrow a = \dfrac{3}{r^3} = \dfrac{3}{3^3} = \dfrac{1}{3^2} = \dfrac{1}{9}.

To find the 7th term, use the formula a_7 = ar^6:

⇒ a_7 = \dfrac{1}{9} \times 3^6

⇒ = \dfrac{1}{9} \times 729

⇒ = 81.

Thus, the 7th term is 81.

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Question 13

Find the geometric progression with fourth term = 54 and seventh term = 1458.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

Given,

∴ a imes r^3 = 54 ……..(i)

Also,

∴ a imes r^6 = 1458 ……..(ii)

Now, divide equation (ii) by equation (i):

\begin{aligned}\Rightarrow \frac{a r^6}{a r^3} = \frac{1458}{54} \\\Rightarrow r^3 = 27 \\\Rightarrow r = \sqrt[3]{27} \\\Rightarrow r = 3.\end{aligned}

Next, substitute the value of r back into equation (i):

∴ a(3)^3 = 54

∴ 27a = 54

∴ a = \frac{54}{27} = 2.

Now, calculate the subsequent terms:

• Second term, a_2 = ar = 2 \times 3 = 6.
• Third term, a_3 = ar^2 = 2 \times 3^2 = 2 \times 9 = 18.

Thus, the G.P. is 2, 6, 18, 54, \ldots

Hence, G.P. = 2, 6, 18, 54, \ldots

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Question 14

Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

From the problem, we have:

⇒ The second term is a_2 = 6
⇒ ar = 6 ……..(i)

Additionally, the fifth term is 9 times the third term:

⇒ a_5 = 9a_3
⇒ ar^4 = 9ar^2

Dividing both sides by ar^2 gives:

⇒ \dfrac{\text{ar}^4}{\text{ar}^2} = 9

⇒ r^2 = 9

⇒ r = \sqrt{9}

⇒ r = ±3

Since all terms of the G.P. are positive, r cannot be -3.

∴ r = 3

Substitute r = 3 back into equation (i):

⇒ 3a = 6

⇒ a = 2

Thus, the G.P. is a, ar, ar^2, ar^3, \ldots

= 2, 6, 18, 54, \ldots

Hence, G.P. = 2, 6, 18, 54, …….

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Question 15

The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

Given:

∴ The fourth term a r^3 = 10 ……..(i)

And,

∴ The seventh term a r^6 = 80 ……..(ii)

To find r, divide equation (ii) by equation (i):

\begin{aligned}\dfrac{a r^6}{a r^3} = \dfrac{80}{10} \\\Rightarrow r^3 = 8 \\\Rightarrow r = \sqrt[3]{8} \\\Rightarrow r = 2.\end{aligned}

Substitute r = 2 back into equation (i):

\begin{aligned}a (2)^3 = 10 \\\Rightarrow 8a = 10 \\\Rightarrow a = \dfrac{10}{8} \\\Rightarrow a = \dfrac{5}{4}.\end{aligned}

Let n represent the number of terms in the G.P. Then, the last term is given by:

a r^{n-1} = 2560

Substitute a = \frac{5}{4} and r = 2:

\begin{aligned}\dfrac{5}{4} \times (2)^{n-1} = 2560 \\\Rightarrow (2)^{n-1} = \dfrac{4}{5} \times 2560 \\\Rightarrow (2)^{n-1} = 4 \times 512 \\\Rightarrow (2)^{n-1} = 2048 \\\Rightarrow (2)^{n-1} = (2)^{11} \\\Rightarrow n-1 = 11 \\\Rightarrow n = 12.\end{aligned}

Thus, the first term is \dfrac{5}{4}, the common ratio is 2, and the number of terms is 12.

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Question 16

If the fourth and ninth terms of a G.P. are 54 and 13122 respectively, find the G.P. Also, find its general term.

Consider the first term of the geometric progression as a and the common ratio as r.

We know:

∴ The fourth term a_4 = 54

⇒ ar^3 = 54 ……..(i)

And also:

∴ The ninth term a_9 = 13122

⇒ ar^8 = 13122 ………(ii)

Now, by dividing equation (ii) by equation (i), we have:

\begin{aligned}\Rightarrow \dfrac{ar^8}{ar^3} = \dfrac{13122}{54} \\\Rightarrow r^5 = 243 \\\Rightarrow r^5 = 3^5 \\\Rightarrow r = 3.\end{aligned}

By substituting r = 3 back into equation (i), we get:

\begin{aligned}\Rightarrow a(3)^3 = 54 \\\Rightarrow 27a = 54 \\\Rightarrow a = 2.\end{aligned}

The general term of a G.P. is given by ar^{n-1}:

= 2(3)^{n-1} = 2 \times 3^{n-1}

Now, let’s find the second and third terms:

The second term of the G.P. is:

2^\text{nd} \text{ term} = 2 \times 3^{2-1} = 2 \times 3 = 6.

The third term of the G.P. is:

3^\text{rd} \text{ term} = 2 \times 3^{3-1} = 2 \times 3^2 = 18.

Thus, the G.P. is 2, 6, 18, 54, \ldots

Hence, G.P. = 2, 6, 18, 54, \ldots\ and general term = 2 \times 3^{n-1}

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Question 17

The fifth, eighth and eleventh terms of a geometric progression are p, q and r respectively. Show that : q^2 = pr.

Consider the first term of the geometric progression as A and the common ratio as R.

Given that:

⇒ The fifth term a_5 = p can be expressed as AR^4 = p ……..(i)

Similarly,

⇒ The eighth term a_8 = q is AR^7 = q ………(ii)

And,

⇒ The eleventh term a_{11} = r is AR^{10} = r ………(iii)

Now, multiply equations (i) and (iii):

AR^4 \times AR^{10} = pr

This simplifies to:

A^2R^{14} = pr ………(iv)

Next, square equation (ii):

(AR^7)^2 = q^2

This gives:

A^2R^{14} = q^2 ………(v)

Notice that the left-hand sides of equations (iv) and (v) are identical, which implies their right-hand sides must be equal as well.

∴ q^2 = pr.

Hence, proved that q^2 = pr.

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Question 18

Find the seventh term from the end of the series :

\sqrt{2}, 2, 2\sqrt{2}, .........., 32

To determine if the sequence is a geometric progression, observe that:

\dfrac{2}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2}.

This confirms the sequence is a G.P. with common ratio r = \sqrt{2}.

Assume there are n terms in the series. We have:

ar^{n - 1} = 32

Substitute the values:

\sqrt{2} \times (\sqrt{2})^{n - 1} = 32

This simplifies to:

\sqrt{2}^{n} = 32

Recognizing that ( 32 = (\sqrt{2})^{10} ), we equate:

(\sqrt{2})^n = (\sqrt{2})^{10}

Thus, n = 10.

The formula for finding the m^{th} term from the end is equivalent to finding the ( (n – m + 1)^{th} ) term from the start.

For the 7th term from the end, calculate the ( (10 – 7 + 1) = 4^{th} ) term from the beginning.

The 4th term a_4 is given by:

a_4 = ar^3

Substitute the known values:

a_4 = \sqrt{2}(\sqrt{2})^3

Simplifying, we get:

a_4 = \sqrt{2} \times 2\sqrt{2} a_4 = 4.

Hence, 7th term from end is 4.

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Question 19

Find the third term from the end of the G.P.

\dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, ........, 162.

To determine the common ratio of the given geometric progression, calculate:

\dfrac{\dfrac{2}{9}}{\dfrac{2}{27}} = \dfrac{2 \times 27}{2 \times 9} = 3.

Thus, the sequence is a G.P. with a common ratio r = 3.

Assume the sequence has n terms. The last term can be represented as:

ar^{n - 1} = 162

Substituting the values, we have:

\dfrac{2}{27} \times (3)^{n - 1} = 162

Rewriting with powers of three:

\dfrac{2}{3^3} \times 3^{n - 1} = 162

This simplifies to:

3^{n - 1 - 3} = \dfrac{162}{2}

Since \dfrac{162}{2} = 81, we have:

3^{n - 4} = 81

Recognizing that 81 is 3^4, we equate:

3^{n - 4} = 3^4

This implies:

n - 4 = 4

Solving gives:

n = 8.

To find the 3rd term from the end, calculate the (n - m + 1)^{th} term from the start. Here, m = 3, so:

(8 - 3 + 1) = 6.

Now, find the 6th term from the start:

a_6 = ar^5

Substituting the values:

= \dfrac{2}{27} \times (3)^5 = \dfrac{2}{27} \times 243

Calculating further:

= 2 \times 9 = 18.

Thus, the 3rd term from the end is 18.

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Question 20

For the G.P. \dfrac{1}{27}, \dfrac{1}{9}, \dfrac{1}{3}, ........., 81;

find the product of fourth term from the beginning and the fourth term from the end.

First, we determine the common ratio of the given geometric progression (G.P.). Calculate it by dividing the second term by the first term:

\dfrac{\dfrac{1}{9}}{\dfrac{1}{27}} = \dfrac{27}{9} = 3.

Thus, the sequence forms a G.P. with a common ratio r = 3.

Suppose there are n terms in this sequence. We know the last term is 81, so:

ar^{n - 1} = 81

Substituting the values, we have:

\dfrac{1}{27} \times (3)^{n - 1} = 81

Multiply both sides by 27:

3^{n - 1} = 81 \times 27

Express 81 and 27 as powers of 3:

3^{n - 1} = 3^4 \times 3^3

Simplifying gives:

3^{n - 1} = 3^7

Equating the exponents, we find:

n - 1 = 7

Thus, n = 8.

Now, calculate the fourth term from the beginning, a_4:

a_4 = ar^3 = \dfrac{1}{27} \times (3)^3

Simplifying, we get:

a_4 = \dfrac{1}{27} \times 27 = 1.

Next, find the fourth term from the end. This is the ((n – 4 + 1))th term from the start, which is the 5th term:

a_5 = ar^4 = \dfrac{1}{27} \times (3)^4

Simplifying, we obtain:

a_5 = \dfrac{1}{27} \times 81 = 3.

Finally, the product of the fourth term from the beginning and the fourth term from the end is:

a_4 \times a_5 = 1 \times 3 = 3.

Hence, the product of the fourth term from the beginning and the fourth term from the end is 3.

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Question 21

If a, b and c are in G.P. and a, x, b, y, c are in A.P. prove that :

(i) \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}

(ii) \dfrac{a}{x} + \dfrac{c}{y} = 2.

Given that a, b, and c form a geometric progression (G.P.), we have:

b^2 = ac \tag{i}

Additionally, a, x, b, y, c form an arithmetic progression (A.P.), leading to:

2x = a + b \quad \text{and} \quad 2y = b + c

Thus, we can express x and y as:

x = \dfrac{a + b}{2} \quad \text{and} \quad y = \dfrac{b + c}{2} \tag{ii}

Part (i): To demonstrate that \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}, substitute the expressions for x and y from (ii):

\text{L.H.S.} = \dfrac{1}{\dfrac{a + b}{2}} + \dfrac{1}{\dfrac{b + c}{2}} = \dfrac{2}{a + b} + \dfrac{2}{b + c}

Combine the fractions:

= \dfrac{2(b + c) + 2(a + b)}{(a + b)(b + c)} = \dfrac{2b + 2c + 2a + 2b}{(a + b)(b + c)}

Simplify by factoring and using (i):

= \dfrac{2(a + c + 2b)}{ab + ac + b^2 + bc} = \dfrac{2(a + c + 2b)}{b(a + b + b + c)} = \dfrac{2(a + c + 2b)}{b(a + c + 2b)} = \dfrac{2}{b} = \text{R.H.S.}

∴ \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b} is proven.

Part (ii): To show \dfrac{a}{x} + \dfrac{c}{y} = 2, use the expressions for x and y from (ii):

\text{L.H.S.} = \dfrac{a}{\dfrac{a + b}{2}} + \dfrac{c}{\dfrac{b + c}{2}} = \dfrac{2a}{a + b} + \dfrac{2c}{b + c}

Factor and simplify:

= 2\left(\dfrac{a}{a + b} + \dfrac{c}{b + c}\right) = 2\left\dfrac{a(b + c) + c(a + b)}{(a + b)(b + c)}\right$$$$

Using (i), simplify further:

= 2\left(\dfrac{ab + ac + ac + bc}{ab + ac + b^2 + bc}\right) = 2 = \text{R.H.S.}

∴ \dfrac{a}{x} + \dfrac{c}{y} = 2 is also proven.

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Question 22

If a, b and c are in A.P. and also in G.P., show that : a = b = c.

Given that a, b, and c form an arithmetic progression (A.P.), we have:

∴ 2b = a + c.

This implies:

b = \dfrac{a + c}{2} …….(i)

Additionally, since a, b, and c are in a geometric progression (G.P.), it follows that:

b^2 = ac.

Substituting the expression for b from equation (i) into the above, we obtain:

\begin{aligned}\Rightarrow \Big(\dfrac{a + c}{2}\Big)^2 = ac \\\Rightarrow \dfrac{a^2 + c^2 + 2ac}{4} = ac \\\Rightarrow a^2 + c^2 + 2ac = 4ac \\\Rightarrow a^2 + c^2 - 2ac = 0 \\\Rightarrow (a - c)^2 = 0 \\\Rightarrow a - c = 0 \\\Rightarrow a = c\end{aligned} …….(ii)

Substituting a = c from equation (ii) back into equation (i), we find:

b = \dfrac{c + c}{2} = \dfrac{2c}{2} = c …….(iii)

From equations (ii) and (iii), it follows that:

a = b = c.

Hence, proved that a = b = c.

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Exercise 11(B)

Question 1(a)

The sum of first four terms of the G.P. 2, 6, 18, ………, is :

• (a) 26
• (b) 80
• (c) 160
• (d) 52

To find the sum of the first four terms of the given geometric progression, we start by determining the common ratio. The common ratio r is calculated as \dfrac{a_{n + 1}}{a_n}, which gives us r = \dfrac{6}{2} = 3.

Now, let’s use the formula for the n-th term of a geometric progression, a_n = ar^{n-1}. For the fourth term, we have:

⇒ a_4 = 2 \times (3)^{4-1}

⇒ a_4 = 2 \times 3^3

⇒ a_4 = 2 \times 27 = 54.

Next, we add the first four terms of the sequence: 2 + 6 + 18 + 54 = 80.

Hence, Option 2 is the correct option.

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Question 1(b)

The 4^th term of a G.P. is 54 and its 7^th term is 1458, the common ratio of this G.P. is :

• (a) \dfrac{1}{3}
• (b) 3
• (c) -3
• (d) -\dfrac{1}{3}

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

The formula for the n^{th} term of a G.P. is given by:

a_n = ar^{n-1}

We know the 4^{th} term is 54, so:

a_4 = 54
ar^{4-1} = 54
ar^3 = 54 ( \text{…(1)} )

Similarly, the 7^{th} term is given as 1458:

a_7 = 1458
ar^{7-1} = 1458
ar^6 = 1458 ( \text{…(2)} )

To find the common ratio, divide equation (2) by equation (1):

\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54}
\Rightarrow r^3 = 27
\Rightarrow r^3 = 3^3
\Rightarrow r = 3

Hence, Option 2 is the correct option.

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Question 1(c)

8, x and 32 are in G.P., then the value of x is :

• (a) 24
• (b) 256
• (c) 40
• (d) 16

We have the numbers 8, x, and 32 forming a geometric progression (G.P.).

∴ \dfrac{x}{8} = \dfrac{32}{x}

This implies:

\Rightarrow x^2 = 8 \times 32

Calculating further:

\Rightarrow x^2 = 256

Taking the square root on both sides, we find:

\Rightarrow x = \sqrt{256} = 16

Hence, Option 4 is the correct option.

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Question 1(d)

The sum of three terms (numbers) of a G.P. is 3\dfrac{1}{2} and their product is 1; the numbers are :

• (a) \dfrac{1}{2}, 1 and 2
• (b) \dfrac{1}{3}, 3 and 9
• (c) 1, \dfrac{1}{2} and 2
• (d) 2, \dfrac{1}{2} and 1

Assume the three terms of the geometric progression are \dfrac{a}{r}, a, ar.

According to the problem, the product of these three terms equals 1.

\begin{aligned}\therefore \dfrac{a}{r} \times a \times ar = 1 \\\Rightarrow a^3 = 1 \\\Rightarrow a^3 = 1^3 \\\Rightarrow a = 1.\end{aligned}

Now, consider that the sum of these terms is 3\dfrac{1}{2}.

\begin{aligned}\Rightarrow \dfrac{a}{r} + a + ar = 3\dfrac{1}{2} \\\Rightarrow \dfrac{1}{r} + 1 + r = \dfrac{7}{2} \quad \end{aligned}\because a = 1\begin{aligned} \\\Rightarrow \dfrac{1 + r + r^2}{r} = \dfrac{7}{2} \\\Rightarrow 2(r^2 + r + 1) = 7r \\\Rightarrow 2r^2 + 2r + 2 = 7r \\\Rightarrow 2r^2 + 2r - 7r + 2 = 0 \\\Rightarrow 2r^2 - 5r + 2 = 0.\end{aligned}

We can factor this quadratic equation as follows:

\begin{aligned}2r^2 - 4r - r + 2 = 0 \\\Rightarrow 2r(r - 2) - 1(r - 2) = 0 \\\Rightarrow (2r - 1)(r - 2) = 0.\end{aligned}

This gives us two possible values for r:

\begin{aligned}2r - 1 = 0 \text{ or } r - 2 = 0 \\\Rightarrow 2r = 1 \text{ or } r = 2 \\\Rightarrow r = \dfrac{1}{2} \text{ or } r = 2.\end{aligned}

Let’s evaluate both possibilities:

1. If r = \dfrac{1}{2}, the terms are:

\dfrac{a}{r}, a, ar \Rightarrow \dfrac{1}{\dfrac{1}{2}}, 1, 1 \times \dfrac{1}{2} \Rightarrow 2, 1, \dfrac{1}{2}.

1. If r = 2, the terms are:

\dfrac{a}{r}, a, ar \Rightarrow \dfrac{1}{2}, 1, 1 \times 2 \Rightarrow \dfrac{1}{2}, 1, 2.

Hence, Option 1 is the correct option.

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Question 1(e)

The sum of 20 terms of the G.P. 10, 20, 40, …… is :

• (a) 10(2^19 – 1)
• (b) 10(2^21 – 1)
• (c) 10(2^20 – 1)
• (d) none of these

To determine the sum of the first 20 terms of the given geometric progression, we start by identifying the common ratio. The sequence provided is 10, 20, 40, and so on.

The first term, a, is 10. The common ratio, r, is calculated as \dfrac{a_{2}}{a_{1}} = \dfrac{20}{10} = 2.

For a geometric progression where |r| > 1, the formula to find the sum of the first n terms (S_n) is:

S_n = \dfrac{a(r^n - 1)}{r - 1}

Applying this formula to our problem:

S_{20} = \dfrac{10(2^{20} - 1)}{2 - 1}

This simplifies to:

S_{20} = \dfrac{10(2^{20} - 1)}{1}

Which further simplifies to:

S_{20} = 10(2^{20} - 1)

Hence, option 3 is the correct option.

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Question 2(i)

Find the sum of G.P. :

1 + 3 + 9 + 27 + …….. to 12 terms

The sequence given is a geometric progression where the first term a is 1 and the common ratio r is calculated as \dfrac{3}{1} = 3.

To find the sum of the first 12 terms of this G.P., we’ll use the formula for the sum of a geometric series:

S = \dfrac{a(r^n - 1)}{r - 1} \quad \text{(since } |r| > 1)

Substituting the known values, we have:

S = \dfrac{1(3^{12} - 1)}{3 - 1}

Calculating further:

= \dfrac{531441 - 1}{2} = \dfrac{531440}{2} = 265720

Thus, the sum of the series to 12 terms is 265720.

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Question 2(ii)

Find the sum of G.P. :

0.3 + 0.03 + 0.003 + 0.0003 + …… to 8 terms.

The common ratio r for this geometric progression can be calculated as \dfrac{0.03}{0.3} = 0.1.

The formula to find the sum of the first n terms of a geometric progression, when |r| < 1, is:

S = \dfrac{a(1 - r^n)}{(1 - r)}

Applying this formula, we have:

S = \dfrac{0.3\Big1 – (0.1)^8\Big}{1 - 0.1}

This simplifies to:

= \dfrac{0.3\Big1 – \Big(\dfrac{1}{10}\Big)^8\Big}{0.9}

Further simplifying, we get:

= \dfrac{1}{3}\Big(1 - \dfrac{1}{10^8}\Big)

Hence, the sum is ( \dfrac{1}{3}\Big(1 – \dfrac{1}{10^8}\Big) ).

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Question 2(iii)

Find the sum of G.P. :

1 - \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} + ....... to 9 terms

To determine the sum of the given geometric progression, first identify the common ratio r. Here, r is calculated as \dfrac{-\dfrac{1}{2}}{1} = -\dfrac{1}{2}.

The formula for the sum of the first n terms of a geometric progression, when |r| < 1, is:

S = \dfrac{a(1 - r^n)}{(1 - r)}

Substitute the values into the formula:

S = \dfrac{1\Big1 – \Big(-\dfrac{1}{2}\Big)^9\Big}{1 - \Big(-\dfrac{1}{2}\Big)}

Simplify the expression:

= \dfrac{\Big1 +\Big(\dfrac{1}{2^9}\Big)\Big}{1 + \dfrac{1}{2}}

Further simplification gives:

= \dfrac{\Big(1 + \dfrac{1}{2^9}\Big)}{\dfrac{3}{2}}

Finally, simplify to:

= \dfrac{2}{3}\Big(1 + \dfrac{1}{2^9}\Big)

Thus, the sum is \dfrac{2}{3}\Big(1 + \dfrac{1}{2^9}\Big).

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Question 3

How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461 ?

Assume we need to add n terms.

Here, the common ratio r is calculated as \dfrac{4}{1} = 4.

∴ The sum of n terms of a geometric progression is given by:

S = \dfrac{a(r^n - 1)}{(r - 1)} \quad \text{(since $|r| > 1$)}

Given S = 5461, substitute the values:

5461 = \dfrac{1[(4)^n - 1]}{4 - 1}

Simplifying, we have:

\dfrac{4^n - 1}{3} = 5461 \Rightarrow 4^n - 1 = 16383 \Rightarrow 4^n = 16384

Recognizing that 16384 is 4^7, we find:

\Rightarrow 4^n = 4^7

\therefore n = 7.

Thus, 7 terms must be added to obtain a sum of 5461.

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Question 4

The first term of a G.P. is 27 and its 8^th term is \dfrac{1}{81}. Find the sum of its first 10 terms.

We know the first term a = 27 and the eighth term a_8 = \dfrac{1}{81}.

∴ ar^7 = \dfrac{1}{81}

⇒ 27r^7 = \dfrac{1}{81}

⇒ r^7 = \dfrac{1}{81 \times 27}

⇒ r^7 = \dfrac{1}{3^4 \times 3^3}

⇒ ( r^7 = \Big(\dfrac{1}{3}\Big)^7 )

⇒ r = \dfrac{1}{3}.

Since r < 1, we use the formula for the sum of the first n terms of a G.P.,

S = \dfrac{a(1 - r^n)}{1 - r}

Substituting the values, we have:

S = \dfrac{27\Big1 – \Big(\dfrac{1}{3}\Big)^{10}\Big}{1 - \dfrac{1}{3}}

This simplifies to:

S = \dfrac{27\Big(1 - \dfrac{1}{3^{10}}\Big)}{\dfrac{2}{3}}

⇒ S = \dfrac{81}{2}\Big(1 - \dfrac{1}{3^{10}}\Big)

Therefore, the sum of the first 10 terms is ( \dfrac{81}{2}\Big(1 – \dfrac{1}{3^{10}}\Big) ).

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Question 5

A boy spends ₹ 10 on first day, ₹ 20 on second day, ₹ 40 on third day and so on. Find how much in all, will he spend in 12 days?

The sequence of spending forms a geometric progression: 10, 20, 40, and so on.

Here, the first term a = 10 and the common ratio r = \dfrac{20}{10} = 2.

To find the total spending over 12 days, we use the sum formula for a geometric progression where |r| > 1:

S = \dfrac{a(r^n - 1)}{r - 1}

Substituting the known values:

S = \dfrac{10(2^{12} - 1)}{2 - 1}

This simplifies to:

S = 10(2^{12} - 1).

Hence, ₹10(2^{12} – 1).

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Question 6

The 4^th and the 7^th terms of a G.P. are \dfrac{1}{27} \text{ and } \dfrac{1}{729} respectively. Find the sum of n terms of this G.P.

We know that the 4th term of the G.P. is \dfrac{1}{27}, which can be expressed as ar^3 = \dfrac{1}{27} ……..(i). Similarly, the 7th term is \dfrac{1}{729}, giving us ar^6 = \dfrac{1}{729} ……..(ii).

By dividing equation (ii) by equation (i), we have:

\begin{aligned}\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{\dfrac{1}{729}}{\dfrac{1}{27}} \\\Rightarrow r^3 = \dfrac{27}{729} \\\Rightarrow r^3 = \dfrac{1}{27} \\\Rightarrow r = \dfrac{1}{3}.\end{aligned}

Plugging the value of r back into equation (i), we find:

\begin{aligned}\Rightarrow a\Big(\dfrac{1}{3}\Big)^3 = \dfrac{1}{27} \\\Rightarrow a \times \dfrac{1}{27} = \dfrac{1}{27} \\\Rightarrow a = 1.\end{aligned}

Given that r < 1, the formula for the sum of n terms, S, is:

\begin{aligned}S = \dfrac{a(1 - r^n)}{(1 - r)} \\= \dfrac{1\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big]}{1 - \dfrac{1}{3}} \\= \dfrac{\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big]}{\dfrac{2}{3}} \\= \dfrac{3}{2}\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big] \\= \dfrac{3}{2}\Big(1 - \dfrac{1}{3^n} \Big)\end{aligned}

Hence, sum = \dfrac{3}{2}\Big(1 - \dfrac{1}{3^n} \Big).

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Question 7

A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.

Consider the last term as the n^{th} term.

∴ ar^{n-1} = 486

This implies:

a(3)^{n-1} = 486

Rearranging gives:

a \cdot 3^n \cdot 3^{-1} = 486

Thus:

3^n = \dfrac{486 \times 3}{a} = \dfrac{1458}{a}

Since the common ratio r > 1, the sum of the series is given by:

S = \dfrac{a(r^n - 1)}{r - 1}

Substituting the known values, we have:

728 = \dfrac{a \times (3^n - 1)}{3 - 1}

This simplifies to:

728 = \dfrac{a \times \Big(\dfrac{1458}{a} - 1\Big)}{2}

Further simplifying gives:

728 = \dfrac{a \times (1458 - a)}{2a}

Cancelling a from numerator and denominator, we get:

\dfrac{1458 - a}{2} = 728

Multiplying both sides by 2 gives:

1458 - a = 1456

Solving for a, we find:

a = 1458 - 1456

Thus, a = 2.

Hence, the first term is 2.

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Question 8

Find the sum of G.P. : 3, 6, 12, ……., 1536.

The common ratio of the given geometric progression is calculated as \dfrac{6}{3} = 2.

Assume that 1536 is the n^{th} term of the sequence.

\begin{aligned}\therefore ar^{n - 1} = 1536 \\\Rightarrow 3 \times (2)^{n - 1} = 1536 \\\Rightarrow (2)^{n - 1} = 512 \\\Rightarrow (2)^{n - 1} = (2)^{9} \\\Rightarrow n - 1 = 9 \\\Rightarrow n = 10.\end{aligned}

Since the common ratio r > 1, we use the formula for the sum of a geometric series:

\begin{aligned}S = \dfrac{a(r^n - 1)}{(r - 1)} \\= \dfrac{3 \times (2^{10} - 1)}{2 - 1} \\= 3(2^{10} - 1) \\= 3(1024 - 1) \\= 3 \times 1023 \\= 3069.\end{aligned}

Hence, sum = 3069.

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Question 9

How many terms of the series 2 + 6 + 18 + …… must be taken to make the sum equal to 728?

The first step is to identify the common ratio of the series. Here, the common ratio r is calculated as \dfrac{6}{2} = 3.

Let’s assume n is the number of terms we need.

Because the common ratio r is greater than 1, we use the formula for the sum of a geometric series:

S = \dfrac{a(r^n - 1)}{(r - 1)}

Substitute the values into the formula:

728 = \dfrac{2 \times (3^n - 1)}{3 - 1}

Simplifying further:

\Rightarrow 728 = \dfrac{2(3^n - 1)}{2}

This reduces to:

\Rightarrow 3^n - 1 = 728

Adding 1 to both sides gives:

\Rightarrow 3^n = 729

Notice that 729 is 3 raised to the power of 6:

\Rightarrow 3^n = 3^6

Thus, n = 6.

Hence, 6 terms must be taken to make the sum equal to 728.

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Question 10

In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.

We know that the ratio of the sum of the first three terms to the sum of the first six terms is given by:

\Rightarrow \dfrac{S_3}{S_6} = \dfrac{125}{152}

For a geometric progression, the sum of the first n terms is expressed as S_n = \dfrac{a(r^n - 1)}{r - 1}. Applying this for S_3 and S_6, we have:

\Rightarrow \dfrac{\dfrac{a(r^3 - 1)}{r - 1}}{\dfrac{a(r^6 - 1)}{r - 1}} = \dfrac{125}{152}

The a(r - 1) terms cancel out, simplifying to:

\Rightarrow \dfrac{r^3 - 1}{r^6 - 1} = \dfrac{125}{152}

Notice that r^6 - 1 can be factored as (r^3 - 1)(r^3 + 1):

\Rightarrow \dfrac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \dfrac{125}{152}

After cancelling r^3 - 1 from both numerator and denominator, we get:

\Rightarrow \dfrac{1}{r^3 + 1} = \dfrac{125}{152}

Taking the reciprocal gives:

\Rightarrow r^3 + 1 = \dfrac{152}{125}

Subtracting 1 from both sides results in:

\Rightarrow r^3 = \dfrac{152}{125} - 1

Simplifying the right side:

\Rightarrow r^3 = \dfrac{152 - 125}{125}

This simplifies further to:

\Rightarrow r^3 = \dfrac{27}{125}

Recognizing the cube, we have:

\Rightarrow r^3 = \Big(\dfrac{3}{5}\Big)^3

Thus, taking the cube root, we find:

\Rightarrow r = \dfrac{3}{5}.

Hence, common ratio = \dfrac{3}{5}.

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Question 11

If the sum of 1 + 2 + 2^2 + …….. + 2^n – 1 is 255, find the value of n.

The common ratio here is r=2.

Given that |r| > 1, we use the formula for the sum of a geometric series:

S = \dfrac{a(r^n - 1)}{(r - 1)}

Substituting the given values and solving for n:

\begin{aligned}255 &= \dfrac{1(2^n - 1)}{(2 - 1)} \\\Rightarrow 2^n - 1 &= 255 \\\Rightarrow 2^n &= 256 \\\Rightarrow 2^n &= 2^8 \\\Rightarrow n &= 8\end{aligned}

Hence, n = 8.

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Question 12

The sum of three numbers in G.P. is \dfrac{39}{10} and their product is 1. Find the numbers.

Consider the numbers as \dfrac{a}{r}, a, ar.

We know that their product is 1.

\therefore \dfrac{a}{r} \times a \times ar = 1

⇒ a^3 = 1

⇒ a = 1.

The sum of the numbers is \dfrac{39}{10}.

\therefore \dfrac{a}{r} + a + ar = \dfrac{39}{10}

Substituting a = 1, we have:

\dfrac{1}{r} + 1 + r = \dfrac{39}{10}

Rearranging gives:

\dfrac{1 + r + r^2}{r} = \dfrac{39}{10}

Cross-multiplying results in:

10(1 + r + r^2) = 39r

Expanding and rearranging terms:

10 + 10r + 10r^2 = 39r 10r^2 - 29r + 10 = 0

Applying factorization:

10r^2 - 25r - 4r + 10 = 0 5r(2r - 5) - 2(2r - 5) = 0 (5r - 2)(2r - 5) = 0

Solving for r gives:

5r - 2 = 0 \text{ or } 2r - 5 = 0

⇒ r = \dfrac{2}{5} or r = \dfrac{5}{2}

For r = \dfrac{2}{5}:

• \dfrac{a}{r} = \dfrac{1}{\dfrac{2}{5}} = \dfrac{5}{2}
• a = 1
• ar = 1 \times \dfrac{2}{5} = \dfrac{2}{5}

For r = \dfrac{5}{2}:

• \dfrac{a}{r} = \dfrac{1}{\dfrac{5}{2}} = \dfrac{2}{5}
• a = 1
• ar = 1 \times \dfrac{5}{2} = \dfrac{5}{2}

Thus, the numbers are \dfrac{5}{2}, 1, \dfrac{2}{5} or \dfrac{2}{5}, 1, \dfrac{5}{2}.

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Question 13

The first term of a G.P. is -3 and the square of the second term is equal to its 4^th term. Find its 7^th term.

We know the first term a = -3. The problem states that the square of the second term equals the fourth term.

This translates mathematically to:
(ar)^2 = ar^3

Substituting the value of a, we have:
(-3r)^2 = -3r^3

Simplifying this equation:
9r^2 = -3r^3

Dividing both sides by r^2 (assuming r \neq 0), we get:
9 = -3r

Solving for r, we find:
r = -3

Now, to find the 7th term a_7, calculate ar^6:
a_7 = (-3)(-3)^6

Calculating further:
= -3 \times 729
= -2187

Hence, 7^\text{th} term = -2187.

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Question 14

Find the 5^th term of the G.P. \dfrac{5}{2}, 1, .......

The common ratio r of the given geometric progression can be calculated as \dfrac{1}{\dfrac{5}{2}} = \dfrac{2}{5}.

To find the 5th term, we’ll use the formula for the n^{th} term of a G.P., which is a_n = ar^{n-1}. So, for the 5th term, we have:

a_5 = ar^4

Substituting the values, we get:

= \dfrac{5}{2} \times \left(\dfrac{2}{5}\right)^4

Calculating further:

= \dfrac{5}{2} \times \dfrac{16}{625}

= \dfrac{8}{125}.

Hence, 5^\text{th} term = \dfrac{8}{125}.

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Question 15

The first two terms of a G.P. are 125 and 25 respectively. Find the 5^th and the 6^th terms of the G.P.

We start with the given terms of a geometric progression:

⇒ First term a = 125

⇒ Second term a_2 = 25

Since a_2 = ar, we have:

⇒ ar = 25

Substituting the value of a:

⇒ 125r = 25

Solving for r, we get:

⇒ r = \dfrac{25}{125} = \dfrac{1}{5}.

To find the 5th term, a_5 = ar^4:

⇒ a_5 = 125 \times \Big(\dfrac{1}{5}\Big)^4

⇒ a_5 = 125 \times \dfrac{1}{625}

⇒ a_5 = \dfrac{1}{5}.

For the 6th term, a_6 = ar^5:

⇒ a_6 = 125 \times \Big(\dfrac{1}{5}\Big)^5

⇒ a_6 = 125 \times \dfrac{1}{3125}

⇒ a_6 = \dfrac{1}{25}.

Hence, a_5 = \dfrac{1}{5} \text{ and } a_6 = \dfrac{1}{25}.

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Question 16

Find the sum of the sequence -\dfrac{1}{3}, 1, -3, 9, ..... upto 8 terms.

To determine the sum of this geometric progression, first identify the common ratio, r. The second term is 1, and the first term is -\dfrac{1}{3}, so the common ratio is calculated as \dfrac{1}{-\dfrac{1}{3}} = -3.

Notice that the absolute value of the common ratio, |r|, is greater than 1. Therefore, we use the formula for the sum of a geometric series:

S = \dfrac{a(r^n - 1)}{(r - 1)}

Substitute the known values: a = -\dfrac{1}{3}, r = -3, and n = 8:

S = \dfrac{-\dfrac{1}{3}[(-3)^8 - 1]}{-3 - 1}

Simplify the expression:

S = -\dfrac{1}{3} \times \dfrac{[(-1)^8(3)^8 - 1]}{-4}

Since (-1)^8 = 1, the expression becomes:

S = \dfrac{1}{12}(3^8 - 1)

Hence, sum = \dfrac{1}{12}(3^8 - 1).

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Question 17

The first term of a G.P. is 27. If the 8^th term be \dfrac{1}{81}, what will be the sum of 10 terms?

We know the first term a of the geometric progression (G.P.) is 27.

For the 8th term, given by a_8 = ar^7, we have:

ar^7 = \dfrac{1}{81}

Substituting a = 27, we get:

27r^7 = \dfrac{1}{81}

Dividing both sides by 27, we find:

r^7 = \dfrac{1}{81 \times 27}

Simplifying the right side, we have:

r^7 = \dfrac{1}{3^7}

This implies:

r^7 = \Big(\dfrac{1}{3}\Big)^7

Thus, r = \dfrac{1}{3}.

Since r < 1, the sum of the first 10 terms S is calculated using the formula:

S = \dfrac{a(1 - r^n)}{(1 - r)}

Substituting the values, we have:

S = \dfrac{27\Big1 – \Big(\dfrac{1}{3}\Big)^{10}\Big}{1 - \dfrac{1}{3}}

Simplifying further:

S = \dfrac{27\Big1 – \Big(\dfrac{1}{3}\Big)^{10}\Big}{\dfrac{2}{3}}

S = \dfrac{81}{2}\Big1 – \Big(\dfrac{1}{3}\Big)^{10}\Big$$$$

Thus, the sum of the first 10 terms is:

\dfrac{81}{2}(1 - 3^{-10})

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Question 18

Find a G.P. for which the sum of first two terms is -4 and the fifth term is 4 times the third term.

We know that the sum of the first two terms of the G.P. is -4.

⇒ a + ar = -4

This can be expressed as:

⇒ a(1 + r) = -4 ……..(i)

It is also given that the fifth term is 4 times the third term:

⇒ ar^4 = 4ar^2

From this, we can deduce:

⇒ r^2 = 4

⇒ r = \pm 2

Let’s first consider r = 2.

Substitute r = 2 into equation (i):

⇒ a(1 + 2) = -4

⇒ 3a = -4

⇒ a = -\dfrac{4}{3}.

Thus, the G.P. is:

-\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}, \ldots

Now, consider r = -2.

Substitute r = -2 into equation (i):

⇒ a(1 + (-2)) = -4

⇒ -a = -4

⇒ a = 4.

Thus, the G.P. is:

4, -8, 16, -32, \ldots

Hence, G.P. = -\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3}, \ldots or 4, -8, 16, -32, \ldots.

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Test Yourself

Question 1(a)

-1, k and -1 are three consecutive terms of a G.P., then

(i) k = 1

(ii) k = -1

Which of the following is valid ?

• (a) only 1
• (b) only 2
• (c) both 1 and 2
• (d) either 1 or -1

Consider the sequence: -1, k, and -1, forming a geometric progression.

∴ We have the relationship:

\dfrac{k}{-1} = \dfrac{-1}{k}

This implies:

k \times k = -1 \times -1

⇒ k^2 = 1

Solving for k, we find:

k = \sqrt{1} = \pm 1

Thus, k can be either 1 or -1.

Hence, Option 4 is the correct option.

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Question 1(b)

x + 9, 10 and 4 are in G.P. The value of x is :

• (a) 16
• (b) 8
• (c) -16
• (d) 0

Consider the terms given: x + 9, 10, and 4, which form a geometric progression (G.P.).

Since these terms are in G.P., the ratio between consecutive terms must be equal. Therefore, we have:

\dfrac{10}{x + 9} = \dfrac{4}{10}

Cross-multiplying gives:

4(x + 9) = 10 \times 10

This simplifies to:

4x + 36 = 100

Subtract 36 from both sides:

4x = 100 - 36

Thus:

4x = 64

Dividing both sides by 4, we find:

x = \dfrac{64}{4} = 16.

Hence, option 1 is the correct option.

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Question 1(c)

The common ratio of a G.P. is 2 and its 6^th term is 48. The first term is :

• (a) \dfrac{3}{2}
• (b) \dfrac{2}{3}
• (c) 1
• (d) 2

We have the common ratio r = 2 for the geometric progression.

The formula for the n^{th} term of a G.P. is given by:

a_n = a \times r^{n-1}

Here, the 6th term is provided as 48.

a_6 = 48

Substituting in the formula, we have:

a \times 2^{6-1} = 48

This simplifies to:

a \times 2^5 = 48 32a = 48

Solving for a, we divide both sides by 32:

a = \frac{48}{32} = \frac{3}{2}

Hence, Option 1 is the correct option.

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Question 1(d)

Three terms are in G.P., whose product is 27. The middle term is :

• (a) -3
• (b) 3
• (c) -3 and 3
• (d) -3 or 3

Consider the three terms of a geometric progression as \dfrac{a}{r}, a, and ar.

We know that their product is given as 27.

\therefore \dfrac{a}{r} \times a \times ar = 27 \

This simplifies to a^3 = 3^3 \

Thus, a = 3.

Hence, Option 2 is the correct option.

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Question 1(e)

The common ratio of a G.P., whose 4^th term is 27 and 6^th term is 243; is :

• (a) 9
• (b) 3
• (c) \dfrac{1}{3}
• (d) \dfrac{1}{9}

Consider the first term of the geometric progression as a and the common ratio as r.

The formula for the n^{th} term of a G.P. is:

a_n = ar^{n-1}

We know that the 4^{th} term is 27. Therefore,

a_4 = 27

This implies:

ar^{4-1} = 27

So, we have:

ar^3 = 27 ……..(1)

Similarly, the 6^{th} term is given as 243:

a_6 = 243

Thus:

ar^{6-1} = 243

Which gives us:

ar^5 = 243 ……..(2)

Now, dividing equation (2) by equation (1), we find:

\Rightarrow \frac{ar^5}{ar^3} = \frac{243}{27}

Simplifying this, we get:

\Rightarrow r^2 = 9

Taking the square root of both sides, we determine:

\Rightarrow r = \sqrt{9} = 3.

Notice that the common ratio r is 3. Therefore, Option 2 is the correct option.

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Question 1(f)

The third term of a G.P. = 18, the product of its first five terms is :

• (a) 18
• (b) 18^5
• (c) 9
• (d) \sqrt{18}

Consider the first five terms of the geometric progression as:

\dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2

We know the third term of the G.P. is given as 18.

∴ a = 18

To find the product of the first five terms, we calculate:

\dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2

This simplifies to a^5.

Substituting the value of a, we get 18^5.

Hence, Option 2 is the correct option.

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Question 1(g)

G.P. : \dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2},..............

Assertion (A): 5^th term of the given G.P. is 1\dfrac{1}{8}.

Reason (R): If for a G.P., the first term is a, the common ratio is r and the number of terms = n, then sum of the first n terms S~n = \dfrac{a(r^n - 1)}{r - 1} for all r.

• (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.

Given the geometric progression: \dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2}, \ldots

The first term, a, is \dfrac{2}{9}.

To find the common ratio r, calculate \dfrac{\dfrac{1}{3}}{\dfrac{2}{9}} = \dfrac{1 \times 9}{2 \times 3} = \dfrac{3}{2}.

To find the 5th term T_5, use the formula T_n = a \cdot r^{n-1}:

T_5 = \dfrac{2}{9} \times \left(\dfrac{3}{2}\right)^{5 - 1} = \dfrac{2}{9} \times \left(\dfrac{3}{2}\right)^4 = \dfrac{2}{9} \times \dfrac{81}{16} = \dfrac{9}{8} = 1\dfrac{1}{8}.

∴ The assertion (A) is correct.

Now, examining the reason (R):

The formula for the sum of the first n terms of a G.P. is S_n = \dfrac{a(r^n - 1)}{r - 1}, applicable for r \neq 1.

Here, the reason (R) is incorrect because it does not mention the condition r \neq 1.

Thus, reason (R) is false.

Hence, option 1 is the correct option.

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Question 1(h)

For a G.P., its fourth term = x, seventh term = y and tenth term = z.

Assertion (A): x, y and z are in G.P.

Reason (R): y^2 = (ar^6)^2 = ar^3 × ar^9 = xz.

• (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.

Consider the first term of the G.P. as a and the common ratio as r.

Using the formula for the n-th term of a G.P., we have:

⇒ T_n = a \cdot r^{n-1}

Given that the fourth term is x, the seventh term is y, and the tenth term is z, we can write:

⇒ a \cdot r^{4-1} = x, a \cdot r^{7-1} = y, and a \cdot r^{10-1} = z

This simplifies to:

⇒ ar^3 = x, ar^6 = y, and ar^9 = z

To verify if x, y, and z form a G.P., the ratio between consecutive terms must be constant.

Calculating the ratio between y and x:

\Rightarrow \dfrac{y}{x} = \dfrac{ar^6}{ar^3} = r^3.

Similarly, the ratio between z and y is:

\Rightarrow \dfrac{z}{y} = \dfrac{ar^9}{ar^6} = r^3.

Since the ratios are equal, x, y, and z indeed form a G.P.

∴ Assertion (A) is true.

Now, consider:

⇒ y^2 = (ar^6)^2

⇒ y^2 = ar^{12}

⇒ y^2 = ar^3 \times ar^9

⇒ y^2 = xz.

∴ Reason (R) is true.

Hence, option 3 is the correct option.

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Question 1(i)

G.P. : = 3 – 6 + 12 – 24 + …………. – 384

Statement (1): Product of 5^th term from the beginning and 5^th term from the end of the G.P. is -1152.

Statement (2): In an G.P. the product of n^th term from the beginning and n^th term from the end is

1^st term + last term

• (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.

We have the geometric progression: 3, -6, 12, -24, …, -384.

Here, the first term a = 3 and the common ratio r = \frac{-6}{3} = -2. The last term a_n = -384.

Using the formula for the n^{th} term of a G.P., T_n = a \cdot r^{n-1}, we set up the equation:

3 \times (-2)^{n-1} = -384

Dividing through by 3, we get:

(-2)^{n-1} = -\frac{384}{3} = -128

This simplifies to:

(-2)^{n-1} = (-2)^7

∴ n-1 = 7 ⇒ n = 8.

Now, for the 5^{th} term from the beginning:

T_5 = 3 \times (-2)^{5-1} = 3 \times (-2)^4 = 3 \times 16 = 48

For the 5^{th} term from the end, which is the 4^{th} term from the beginning:

T_4 = 3 \times (-2)^{4-1} = 3 \times (-2)^3 = 3 \times (-8) = -24

The product of the 5^{th} term from the beginning and the 5^{th} term from the end is:

48 \times (-24) = -1152

Thus, Statement 1 is correct.

For Statement 2, consider a G.P. with first term a and N total terms. The n^{th} term from the beginning is:

T_n = a \cdot r^{n-1} \quad \text{(1)}

The n^{th} term from the end is:

T_{N-n+1} = a \cdot r^{N-n} \quad \text{(2)}

Multiplying (1) and (2), we have:

T_n \times T_{N-n+1} = a^2 \cdot r^{n-1 + N-n} = a^2 \cdot r^{N-1} \quad \text{(3)}

Now, the sum of the first term and the last term is:

a + a \cdot r^{N-1} = a(1 + r^{N-1}) \quad \text{(4)}

Comparing equations (3) and (4), we see that the product of the n^{th} terms from the beginning and end is not equal to the sum of the first and last terms.

Therefore, Statement 2 is false.

Hence, option 3 is the correct option.

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Question 1(j)

In a G.P., common ratio = 2, first term = 3 and last term = 96.

Statement (1): The number of terms in this G.P. = 96 – 3.

Statement (2): a : ar^n – 1 = 3 : 96

• (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.

We have the first term a = 3, the common ratio r = 2, and the last term as 96.

Using the formula for the n-th term of a geometric progression:

T_n = a \cdot r^{n-1}

Substituting the given values:

3 \times 2^{n-1} = 96

Dividing both sides by 3:

2^{n-1} = \frac{96}{3}

This simplifies to:

2^{n-1} = 32

Recognizing 32 as a power of 2, we write:

2^{n-1} = 2^5

Equating the exponents gives:

n-1 = 5

Thus, n = 5 + 1 = 6. Therefore, there are 6 terms in this G.P.

Now, according to statement 1, the number of terms is calculated as 96 - 3 = 93, which is incorrect.

∴ Statement 1 is false.

Next, consider statement 2:

\frac{a}{ar^{n-1}} = \frac{\text{first term}}{\text{last term}}

Substituting the values:

\frac{3}{96} \rightarrow 3 : 96

This ratio holds true.

∴ Statement 2 is true.

Hence, option 4 is the correct option.

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Question 2

The 5^th and the 8^th terms of a G.P. are 32 and 256 respectively. Find its first term and the common ratio.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

The formula for the n^{th} term of a G.P. is:

⇒ a_n = ar^{n-1}

For the 5^{th} term:

Given a_5 = 32, we have:

⇒ ar^{5-1} = 32

⇒ ar^4 = 32 ……….(1)

For the 8^{th} term:

Given a_8 = 256, we have:

⇒ ar^{8-1} = 256

⇒ ar^7 = 256 ……….(2)

Now, divide equation (2) by equation (1):

\Rightarrow \dfrac{ar^7}{ar^4} = \dfrac{256}{32} \Rightarrow r^3 = 8 \Rightarrow r^3 = 2^3 \Rightarrow r = 2.

Substitute r = 2 back into equation (1):

⇒ a \times (2)^4 = 32

⇒ 16a = 32

⇒ a = \dfrac{32}{16} = 2.

Therefore, the first term is 2 and the common ratio is 2.

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Question 3

The third term of a G.P. is greater than its first term by 9 whereas its second term is greater than the fourth term by 18. Find the G.P.

Assume the first term of the geometric progression (G.P.) is a and the common ratio is r.

According to the formula for the n^{th} term of a G.P., a_n = ar^{n-1}.

Given:

The third term exceeds the first term by 9.

∴ a_3 - a = 9

⇒ ar^2 - a = 9

⇒ a(r^2 - 1) = 9

⇒ a = \dfrac{9}{r^2 - 1} …….(1)

Additionally, the second term is 18 more than the fourth term.

∴ a_2 - a_4 = 18

⇒ ar - ar^3 = 18

⇒ ar(1 - r^2) = 18

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

\begin{aligned}\Rightarrow \dfrac{9}{r^2 - 1} \times r(1 - r^2) = 18 \\\Rightarrow \dfrac{9}{r^2 - 1} \times -r(r^2 - 1) = 18 \\\Rightarrow -9r = 18 \\\Rightarrow r = -\dfrac{18}{9} = -2.\end{aligned}

Substitute r = -2 back into equation (1):

\begin{aligned}\Rightarrow a = \dfrac{9}{(-2)^2 - 1} \\= \dfrac{9}{4 - 1} \\= \dfrac{9}{3} \\= 3.\end{aligned}

Thus, the G.P. is a, ar, ar^2, ar^3, \ldots

= 3, 3 \times -2, 3 \times (-2)^2, 3 \times (-2)^3, \ldots

= 3, -6, 3 \times 4, 3 \times -8, \ldots

= 3, -6, 12, -24, \ldots

Hence, required G.P. = 3, -6, 12, -24, \ldots

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Question 4

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P. Find the value of G.

We have the sequence:

x, 2x + 2, 3x + 3, G as four terms of a geometric progression (G.P.).

\therefore We can write:

\dfrac{2x + 2}{x} = \dfrac{3x + 3}{2x + 2}

\Rightarrow Cross-multiplying gives:

(2x + 2)^2 = x(3x + 3)

On expanding and simplifying, we get:

(2x)^2 + 2^2 + 2 \times 2x \times 2 = 3x^2 + 3x \Rightarrow 4x^2 + 4 + 8x = 3x^2 + 3x

\Rightarrow Rearranging terms:

4x^2 - 3x^2 + 8x - 3x + 4 = 0 \Rightarrow x^2 + 5x + 4 = 0

Using factorization:

x^2 + 4x + x + 4 = 0 \Rightarrow x(x + 4) + 1(x + 4) = 0 \Rightarrow (x + 1)(x + 4) = 0

Thus, we find:

x + 1 = 0 \text{ or } x + 4 = 0 \Rightarrow x = -1 \text{ or } x = -4.

First, substituting x = -1 gives us the terms:

-1, 2(-1) + 2, 3(-1) + 3, G

= -1, -2 + 2, -3 + 3, G

= -1, 0, 0, G

This sequence doesn’t work as a G.P. since the common ratio isn’t consistent.

Now, substituting x = -4 gives:

-4, 2(-4) + 2, 3(-4) + 3, G

= -4, -8 + 2, -12 + 3, G

= -4, -6, -9, G

Here, the common ratio is:

\dfrac{-6}{-4} = \dfrac{3}{2}

Thus, the fourth term G is calculated as:

G = -9 \times \dfrac{3}{2} = -\dfrac{27}{2}

Hence, G = -\dfrac{27}{2}.

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Question 5

The third term of a G.P. is 2. Find the product of the first five terms of this G.P.

Consider the first five terms of the geometric progression as \dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2.

We know that the third term is given as 2.

∴ a = 2

The product of these terms is calculated by multiplying them all together:

\dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2 = a^5

Substituting the value of a:

2^5

This simplifies to 32.

Thus, the product of the first five terms of this G.P. is 32.

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Question 6

The 10^th, 16^th and 22^nd terms of a G.P. are x, y and z respectively. Show that x, y and z are in G.P.

Consider the first term of the geometric progression (G.P.) to be a and the common ratio to be r. According to the formula for the n^{th} term of a G.P., we have:

a_n = ar^{n-1}

Given that the 10^{th}, 16^{th}, and 22^{nd} terms are x, y, and z respectively, we can write:

For the 10^{th} term:
a_{10} = x
x = ar^{10-1}
x = ar^9 \quad \text{...(1)}

For the 16^{th} term:
a_{16} = y
y = ar^{16-1}
y = ar^{15} \quad \text{...(2)}

For the 22^{nd} term:
a_{22} = z
z = ar^{22-1}
z = ar^{21} \quad \text{...(3)}

Now, dividing equation (2) by equation (1):
\begin{aligned}\Rightarrow \dfrac{y}{x} = \dfrac{ar^{15}}{ar^9} \\\Rightarrow \dfrac{y}{x} = \dfrac{r^{15}}{r^9} \\\Rightarrow \dfrac{y}{x} = r^{15-9} \\\Rightarrow \dfrac{y}{x} = r^6.\end{aligned}

Next, dividing equation (3) by equation (2):
\begin{aligned}\Rightarrow \dfrac{z}{y} = \dfrac{ar^{21}}{ar^{15}} \\\Rightarrow \dfrac{z}{y} = \dfrac{r^{21}}{r^{15}} \\\Rightarrow \dfrac{z}{y} = r^{21-15} \\\Rightarrow \dfrac{z}{y} = r^6.\end{aligned}

Notice that \dfrac{y}{x} = \dfrac{z}{y} = r^6.

Hence, proved that x, y, and z are in G.P.

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Question 7

Which term of the G.P. 2, 2\sqrt{2}, 4, ........ \text{ is } 128\sqrt{2} ?

Consider the geometric progression given as 2, 2\sqrt{2}, 4, \ldots

Here, the initial term a is 2.

The common ratio r can be calculated as \frac{2\sqrt{2}}{2} = \sqrt{2}.

We need to find which term of this sequence is 128\sqrt{2}.

We know the formula for the n^{th} term of a G.P. is ar^{n-1}. Set this equal to 128\sqrt{2}:

\Rightarrow 2 \times (\sqrt{2})^{n - 1} = 128\sqrt{2}

Express \sqrt{2} as (2^{\frac{1}{2}}):

\Rightarrow 2 \times (2^{\frac{1}{2}})^{n - 1} = 2^7.\sqrt{2}

Rewriting gives:

\Rightarrow 2 \times (2)^{\frac{n - 1}{2}} = 2^7.2^{\frac{1}{2}}

Now, equate the exponents of 2:

\Rightarrow (2)^{1 + \frac{n - 1}{2}} = (2)^{7 + \frac{1}{2}}

Simplify the powers:

\Rightarrow (2)^{\frac{n - 1 + 2}{2}} = (2)^{\frac{14 + 1}{2}} \Rightarrow (2)^{\frac{n + 1}{2}} = (2)^{\frac{15}{2}}

Since the bases are the same, equate the exponents:

\Rightarrow \frac{n + 1}{2} = \frac{15}{2}

Solving for n gives:

\Rightarrow n + 1 = 15 \Rightarrow n = 15 - 1 = 14.

Thus, the 14th term of the G.P. is 128\sqrt{2}.

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Question 8

Find the 8^th term of a G.P., if its common ratio is 2 and 10^th term is 768.

We know that the common ratio r is 2.

Using the formula for the n-th term of a G.P., we have:

a_n = ar^{n-1}

It’s given that the 10th term of the G.P. is 768, so:

a_{10} = 768

This implies:

ar^{10-1} = 768 ar^9 = 768

Substituting the value of r:

a \times 2^9 = 768 a \times 512 = 768

Solving for a:

a = \dfrac{768}{512} = \dfrac{3}{2}

To find the 8th term, we use:

a_8 = ar^{8-1} a_8 = ar^7

Substituting the values of a and r:

a_8 = \dfrac{3}{2} \times 2^7 a_8 = 3 \times 2^6 a_8 = 3 \times 64 = 192

Therefore, the 8th term of the G.P. is 192.

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Question 9

In a G.P., the 4^th term is 48 and 7^th term is 384. Find its 6^th term.

Consider the first term of the geometric progression to be a and the common ratio to be r.

The formula for the n^{th} term of a G.P. is given by:

⇒ a_n = ar^{n-1}

We know the 4th term is 48:

⇒ a_4 = 48

⇒ ar^{4-1} = 48

⇒ ar^3 = 48 ……….(1)

Also, the 7th term is 384:

⇒ a_7 = 384

⇒ ar^{7-1} = 384

⇒ ar^6 = 384 ……….(2)

By dividing equation (2) by equation (1), we obtain:

\begin{aligned}\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{384}{48} \\\Rightarrow r^3 = 8 \\\Rightarrow r^3 = 2^3 \\\Rightarrow r = 2.\end{aligned}

Now, substitute r = 2 back into equation (1):

⇒ a(2)^3 = 48

⇒ a \times 8 = 48

⇒ a = \dfrac{48}{8} = 6.

To find the 6th term, use the formula:

⇒ a_6 = ar^{6-1}

⇒ a_6 = ar^5

⇒ a_6 = 6 \times 2^5

⇒ a_6 = 6 \times 32

⇒ a_6 = 192.

Therefore, the 6th term of the G.P. is 192.

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Question 10

-\dfrac{5}{3}, x \text{ and } -\dfrac{3}{5} are three consecutive terms of a G.P. Find the value(s) of x.

We have the terms -\dfrac{5}{3}, x, and -\dfrac{3}{5} in a geometric progression.

∴ The common ratio between consecutive terms is the same:

\dfrac{x}{-\dfrac{5}{3}} = \dfrac{-\dfrac{3}{5}}{x}

This implies:

x^2 = -\dfrac{3}{5} \times -\dfrac{5}{3}

Simplifying the right-hand side gives:

x^2 = 1

Taking the square root on both sides, we find:

x = \sqrt{1} = \pm 1

Thus, the values of x are +1 or -1.

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Question 11

Rohit writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instructions that they move the chain similarly. Assuming that the chain is not broken and that it costs ₹ 10 to mail one letter, determine the amount spent on postage when the fourth set of letters is mailed.

Rohit begins by sending a letter to four friends. Each friend then replicates the letter and sends it to four other individuals, continuing the chain in the same manner. This sequence forms a geometric progression (G.P.) as follows: 4, 16, 64, and so on.

In this G.P., the first term a is 4 and the common ratio r is also 4.

To find the total number of letters sent by the fourth set, we use the formula for the sum of a geometric progression:

Sum of G.P. = \dfrac{a(r^n - 1)}{(r - 1)}

Substituting the values, we have:

\begin{aligned}S_4 = \dfrac{4 \times (4^4 - 1)}{4 - 1} \\= \dfrac{4 \times (256 - 1)}{3} \\= \dfrac{4 \times 255}{3} \\= 4 \times 85 \\= 340.\end{aligned}

Therefore, a total of 340 letters are mailed by the time the fourth set is completed.

Since mailing one letter costs ₹ 10, the total expenditure on postage is calculated as follows:

Total cost = Number of letters × ₹ 10 = 340 × ₹ 10 = ₹ 3,400.

Hence, the amount spent on postage is ₹ 3,400.

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Question 12

The fourth term of a G.P. is eight times its seventh term. The fifth term of then G.P. is \dfrac{3}{16}, then find its 12^th term.

To solve this, we use the formula for the n^{th} term of a geometric progression (G.P.):

a_n = ar^{n-1}

It’s given that the fourth term is eight times the seventh term:

a_4 = 8a_7

Substituting into the formula, we have:

ar^{4-1} = 8ar^{7-1}

This simplifies to:

ar^3 = 8ar^6

Cancelling out the common terms, we get:

\dfrac{r^6}{r^3} = \dfrac{a}{8a}

Thus:

r^3 = \dfrac{1}{8}

Taking the cube root, we find:

r = \sqrt[3]{\dfrac{1}{8}} = \dfrac{1}{2}

Next, we know the fifth term is \dfrac{3}{16}:

ar^4 = \dfrac{3}{16}

Substituting the value of r:

a \times \left(\dfrac{1}{2}\right)^4 = \dfrac{3}{16}

This simplifies to:

a \times \dfrac{1}{16} = \dfrac{3}{16}

Solving for a:

a = 3

Now, to find the 12th term:

a_{12} = ar^{12-1} = ar^{11}

Substituting the values of a and r:

= 3 \times \left(\dfrac{1}{2}\right)^{11}

Hence, 12^{th} term = ( 3 \times \left(\dfrac{1}{2}\right)^{11} ).

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Question 13

Richard borrows ₹ 10,000 from his friend Ramesh. Ramesh asks him to return the money just by repaying ₹ 10 at first day and double the amount in the subsequent days than the each previous day continuously for 10 days. Richard thanks Ramesh for his help. Determine whether Ramesh was in loss or profit.

The repayment sequence follows a geometric progression:

10 + 20 + 40 + … up to 10 terms.

Here, the sequence is a geometric progression (G.P.) where the first term ( (a) ) is 10, and the common ratio ( (r) ) is 2.

To find the sum of the first 10 terms, we use the formula for the sum of a G.P.:

S_n = \dfrac{a(r^n - 1)}{(r - 1)}

Plugging in the values, we have:

\Rightarrow S_{10} = \dfrac{10 \times (2^{10} - 1)}{2 - 1}
= 10 \times (1024 - 1)
= 10 \times 1023
= 10,230.

Notice that Ramesh initially lent ₹ 10,000 but received ₹ 10,230 in return. ∴, he gains a profit of ₹ 230.

Hence, Ramesh makes a profit of ₹ 230.

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Case Study Based Question :

Question 1

Case study:
The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. A total of 27 flags have to be fixed at intervals of every 3 m. All the flags are kept at the position of the middle most flag. Rohan was given the responsibility of placing the flags on both the sides of the middle most flag. Rohan could carry only one flag at a time. So he carries one flag from the middle most point, places the flag at the designated place and comes back to the middle most point. He continues this for all the flags on both sides.

Based on the above information, answer the following questions:

(i) The distance covered by Rohan in placing the flags on one side of the middle most flag forms a sequence. Identify the type of sequence by writing at least the first 4 terms of the sequence.

(ii) Find the distance covered by Rohan in placing the 10^th flag and coming back to the middle most flag.

(iii) Find the total distance travelled by Rohan in fixing all the 27 flags.

(i) Since there are 27 flags, the middle flag is the 14th one. Therefore, there are 13 flags on each side of this central flag.

For the first flag on either side, Rohan covers 3 meters to place it and another 3 meters to return, totaling 6 meters. For the second flag, he covers 6 meters each way, making it 12 meters in total.

The distances for the first four flags on one side are:

• 1st flag: 2 \times 3 = 6 m
• 2nd flag: 2 \times 6 = 12 m
• 3rd flag: 2 \times 9 = 18 m
• 4th flag: 2 \times 12 = 24 m

This sequence is an arithmetic progression (A.P.) with a common difference of 6.

Hence, the required sequence is an A.P.

(ii) To find the distance covered when placing the 10th flag and returning, use the formula for the nth term of an A.P.:

a_n = a + (n - 1)d

For the 10th flag, a_{10} = 6 + 6(10 - 1) = 6 + 54 = 60 m.

Hence, the distance covered by Rohan in placing the 10th flag and returning back to the middle most flag = 60 m.

(iii) To calculate the total distance Rohan travels for placing all 13 flags on one side, use the sum formula for an A.P.:

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

For 13 flags on one side:

\begin{aligned}S_{13} = \dfrac{13}{2}[2 \times 6 + (13 - 1) \times 6] \\= \dfrac{13}{2} \times [12 + 12 \times 6] \\= \dfrac{13}{2} \times [12 + 72] \\= \dfrac{13}{2} \times 84 \\= 13 \times 42 \\= 546 \text{ m}.\end{aligned}

Thus, the distance covered on one side is 546 m. Since there are two sides, the total distance is 2 \times 546 = 1092 m.

Hence, total distance travelled by Rohan in fixing all the 27 flags = 1092 m.

Frequently Asked Questions

What is the main difference between an Arithmetic Progression (A.P.) and a Geometric Progression (G.P.)?

The key difference lies in how terms are generated. In an A.P., each term is found by adding a constant 'common difference' to the previous term. In a G.P., each term is found by multiplying the previous term by a constant 'common ratio'.

How do you find the common ratio (r) in a Geometric Progression?

To find the common ratio in a G.P., you simply divide any term by its preceding term. For example, if you have the terms a₁, a₂, a₃, the common ratio 'r' would be a₂/a₁ or a₃/a₂. This ratio must be constant throughout the sequence.

What is the formula for the nth term of a G.P.?

The formula to find the nth term (aₙ) of a Geometric Progression is aₙ = arⁿ⁻¹, where 'a' is the first term, 'r' is the common ratio, and 'n' is the term number you want to find. This formula is fundamental for solving most problems in the chapter.

Are these Selina solutions enough for the ICSE Class 10 board exam?

Yes, these solutions for the Selina Concise Mathematics textbook cover all the concepts and question types for the Geometric Progression chapter. They are structured according to the ICSE curriculum and provide a thorough foundation for your board exam preparation.

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