ICSE Class 10 Maths AP Solutions: Selina Chapter 10
Concise Mathematics Selina Solutions Class 10 ICSE Chapter 10 Arithmetic Progression
ICSE Class 10 Maths Chapter 10 Arithmetic Progression teaches how to recognise an A.P., find its nth term, count the number of terms, calculate sums, and insert arithmetic means. The solutions below use school-style working so that you can see the formula, substitution, calculation, and final answer clearly.
This replacement page is written for Selina Concise Mathematics Class 10 Chapter 10 revision. It keeps the focus on the methods students actually use in Arithmetic Progression: common difference, general term, last term, sum of terms, and checking whether a number belongs to a sequence.
Arithmetic Progression formulas for ICSE Class 10 Maths
An arithmetic progression is a sequence in which the difference between two consecutive terms is constant. If the first term is a, the common difference is d, and the nth term is t_n, then the basic formulas are as follows.
| Use case | Formula | What each symbol means |
|---|---|---|
| Common difference | d=t_2-t_1=t_3-t_2 | Subtract a term from the next term. |
| nth term | t_n=a+(n-1)d | a is the first term; d is the common difference. |
| Number of terms when last term is known | l=a+(n-1)d | Put the last term as l and solve for n. |
| Sum of first n terms | S_n=\frac{n}{2}\{2a+(n-1)d\} | Use this when a, d, and n are known. |
| Sum using first and last term | S_n=\frac{n}{2}(a+l) | Use this when a, l, and n are known. |
| Arithmetic mean between x and y | \frac{x+y}{2} | The middle term of a three-term A.P. |
In this chapter, the same formula may appear in different forms. For example, t_n, a_n, and T_n usually mean the nth term. Keep the symbol consistent in your own solution.
Concept snapshot for Arithmetic Progression
Think of an arithmetic progression like climbing a staircase where every step has the same height. The first step is a. The height added each time is d. To reach the nth step, you do not add d exactly n times; you add it n-1 times because the first term is already there. This is why the formula is t_n=a+(n-1)d, not a+nd.
Worked examples before the Selina solutions
These original examples cover the main question types in Concise Mathematics Selina Solutions Class 10 ICSE Chapter 10 Arithmetic Progression. Use them as a method check before reading the exercise solutions.
Worked Example 1: Find the 20th term of an A.P.
Question: Find the 20th term of the A.P. 6, 11, 16, 21, \ldots.
Step 1: Identify the first term and common difference: a=6 and d=11-6=5.
Step 2: Use the nth term formula t_n=a+(n-1)d.
t_{20}=6+(20-1)\times 5
=6+19\times 5=6+95=101
Final answer: t_{20}=101.
Worked Example 2: Find the sum of an A.P.
Question: Find the sum of the first 15 terms of the A.P. 3, 7, 11, 15, \ldots.
Step 1: Here a=3, d=7-3=4, and n=15.
Step 2: Use S_n=\frac{n}{2}\{2a+(n-1)d\}.
S_{15}=\frac{15}{2}\{2(3)+(15-1)4\}
=\frac{15}{2}\{6+56\}=\frac{15}{2}\times 62=15\times 31=465
Final answer: S_{15}=465.
Worked Example 3: Insert arithmetic means
Question: Insert three arithmetic means between 8 and 28.
Step 1: Three means between 8 and 28 make five terms in all: 8, A, B, C, 28.
Step 2: Use t_5=a+4d, where a=8 and t_5=28.
28=8+4d
4d=20 \quad \Rightarrow \quad d=5
Step 3: Add 5 successively to form the sequence.
8,\ 13,\ 18,\ 23,\ 28
Final answer: The three arithmetic means are 13, 18, 23.
Exercise 10(A): Arithmetic Progression solutions
The following Selina-style solutions cover the foundation questions on identifying an A.P., finding a term, and solving for the number of terms. Each solution uses MathJax so that the calculation is clear.
Question 1: Which of the following sequences are in arithmetic progression?
Step 1: For 2, 6, 10, 14, \ldots, compare consecutive differences.
6-2=4,\quad 10-6=4,\quad 14-10=4
Answer for (i): It is an A.P. because the common difference is 4.
Step 2: For 15, 12, 9, 6, \ldots, compare consecutive differences.
12-15=-3,\quad 9-12=-3,\quad 6-9=-3
Answer for (ii): It is an A.P. because the common difference is -3.
Step 3: For 5, 9, 12, 18, \ldots, compare the first two differences.
9-5=4,\quad 12-9=3
Answer for (iii): It is not an A.P. because the consecutive differences are not equal.
Step 4: For \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots, compare the first two differences.
\frac{1}{3}-\frac{1}{2}=-\frac{1}{6},\quad \frac{1}{4}-\frac{1}{3}=-\frac{1}{12}
Answer for (iv): It is not an A.P. because -\frac{1}{6}\ne -\frac{1}{12}.
Question 2: The nth term of a sequence is 2n-3. Find its fifteenth term.
Step 1: The given term formula is t_n=2n-3.
Step 2: Put n=15.
t_{15}=2(15)-3=30-3=27
Final answer: The fifteenth term is 27.
Question 3: If the pth term of an A.P. is 2p+3, find the A.P.
Step 1: The given term is t_p=2p+3.
Step 2: Put p=1,2,3 to get the first three terms.
t_1=2(1)+3=5
t_2=2(2)+3=7
t_3=2(3)+3=9
Final answer: The A.P. is 5, 7, 9, \ldots.
Question 4: Find the 24th term of 12, 10, 8, 6, \ldots
Step 1: Here a=12 and d=10-12=-2.
Step 2: Use t_n=a+(n-1)d.
t_{24}=12+(24-1)(-2)
=12+23(-2)=12-46=-34
Final answer: The 24th term is -34.
Question 5: Find the 30th term of \frac{1}{2}, 1, \frac{3}{2}, \ldots
Step 1: Here a=\frac{1}{2} and d=1-\frac{1}{2}=\frac{1}{2}.
Step 2: Apply the nth term formula.
t_{30}=\frac{1}{2}+(30-1)\frac{1}{2}
=\frac{1}{2}+\frac{29}{2}=\frac{30}{2}=15
Final answer: The 30th term is 15.
Question 6: Find the 100th term of \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, \ldots
Step 1: Here a=\sqrt{3} and d=2\sqrt{3}-\sqrt{3}=\sqrt{3}.
Step 2: Use t_n=a+(n-1)d.
t_{100}=\sqrt{3}+(100-1)\sqrt{3}
=\sqrt{3}+99\sqrt{3}=100\sqrt{3}
Final answer: The 100th term is 100\sqrt{3}.
Question 7: Find the 50th term of \frac{1}{n}, \frac{n+1}{n}, \frac{2n+1}{n}, \ldots
Step 1: The letter n is part of the terms, so use another letter, say r, for the term position.
Step 2: Find the common difference.
\frac{n+1}{n}-\frac{1}{n}=\frac{n}{n}=1
\frac{2n+1}{n}-\frac{n+1}{n}=\frac{n}{n}=1
Step 3: Thus a=\frac{1}{n} and d=1. Put r=50.
t_{50}=\frac{1}{n}+(50-1)(1)=\frac{1}{n}+49
Final answer: The 50th term is \frac{1}{n}+49.
Question 8: Is 402 a term of 8, 13, 18, 23, \ldots?
Step 1: Here a=8 and d=13-8=5.
Step 2: If 402 is a term, then 402=8+(n-1)5.
402=8+5n-5
402=5n+3
399=5n \quad \Rightarrow \quad n=\frac{399}{5}
Step 3: A term number must be a positive integer. Here \frac{399}{5} is not an integer.
Final answer: 402 is not a term of the given A.P.
Question 9: Find d and the 99th term of 7\frac{3}{4}, 9\frac{1}{2}, 11\frac{1}{4}, \ldots
Step 1: Convert the mixed numbers into improper fractions.
7\frac{3}{4}=\frac{31}{4},\quad 9\frac{1}{2}=\frac{19}{2},\quad 11\frac{1}{4}=\frac{45}{4}
Step 2: Find the common difference.
d=\frac{19}{2}-\frac{31}{4}=\frac{38-31}{4}=\frac{7}{4}
Step 3: Use t_n=a+(n-1)d, with a=\frac{31}{4}, d=\frac{7}{4}, and n=99.
t_{99}=\frac{31}{4}+(99-1)\frac{7}{4}
=\frac{31}{4}+\frac{686}{4}=\frac{717}{4}=179\frac{1}{4}
Final answer: d=\frac{7}{4} and t_{99}=179\frac{1}{4}.
Question 10(i): How many terms are in 4, 7, 10, 13, \ldots, 148?
Step 1: Here a=4, d=3, and the last term is 148.
Step 2: Use l=a+(n-1)d.
148=4+(n-1)3
148=4+3n-3=3n+1
147=3n \quad \Rightarrow \quad n=49
Final answer: The series has 49 terms.
Question 10(ii): How many terms are in 0.5, 0.53, 0.56, \ldots, 1.1?
Step 1: Here a=0.5, d=0.53-0.5=0.03, and l=1.1.
Step 2: Use the last-term formula.
1.1=0.5+(n-1)(0.03)
1.1=0.5+0.03n-0.03=0.47+0.03n
0.63=0.03n \quad \Rightarrow \quad n=21
Final answer: The series has 21 terms.
Question 10(iii): How many terms are in \frac{3}{4}, 1, 1\frac{1}{4}, \ldots, 3?
Step 1: Here a=\frac{3}{4}, d=\frac{1}{4}, and l=3.
Step 2: Use l=a+(n-1)d.
3=\frac{3}{4}+(n-1)\frac{1}{4}
12=3+n-1
12=n+2 \quad \Rightarrow \quad n=10
Final answer: The series has 10 terms.
Question 11: Which term of 1, 4, 7, 10, \ldots is 52?
Step 1: Here a=1, d=3, and t_n=52.
52=1+(n-1)3
52=3n-2 \quad \Rightarrow \quad 54=3n \quad \Rightarrow \quad n=18
Final answer: 52 is the 18th term.
Question 12: If the 5th and 6th terms are 6 and 5, find the 11th term.
Step 1: Use t_n=a+(n-1)d.
a+4d=6 \quad \text{and} \quad a+5d=5
Step 2: Subtract the first equation from the second.
d=5-6=-1
Step 3: Put d=-1 in a+4d=6.
a+4(-1)=6 \quad \Rightarrow \quad a=10
Step 4: Find t_{11}.
t_{11}=10+(11-1)(-1)=10-10=0
Final answer: The 11th term is 0.
Question 13: If t_2+t_5-t_3=10 and t_2+t_9=17, find a and d.
Step 1: Write each term using t_n=a+(n-1)d.
t_2=a+d,\quad t_5=a+4d,\quad t_3=a+2d,\quad t_9=a+8d
Step 2: Use t_2+t_5-t_3=10.
(a+d)+(a+4d)-(a+2d)=10
a+3d=10 \quad \text{...(i)}
Step 3: Use t_2+t_9=17.
(a+d)+(a+8d)=17
2a+9d=17 \quad \text{...(ii)}
Step 4: Multiply equation (i) by 2 and subtract from equation (ii).
(2a+9d)-(2a+6d)=17-20
3d=-3 \quad \Rightarrow \quad d=-1
Step 5: Put d=-1 in a+3d=10.
a+3(-1)=10 \quad \Rightarrow \quad a=13
Final answer: First term a=13 and common difference d=-1.
Question 14: Find the 10th term from the end of 4, 9, 14, \ldots, 254
Step 1: Here a=4, d=5, and l=254. First find the total number of terms.
254=4+(n-1)5
254=5n-1 \quad \Rightarrow \quad 255=5n \quad \Rightarrow \quad n=51
Step 2: The 10th term from the end is the (51-10+1)th term from the beginning.
51-10+1=42
Step 3: Find t_{42}.
t_{42}=4+(42-1)5=4+205=209
Final answer: The 10th term from the end is 209.
Question 15: Determine the A.P. whose 3rd term is 5 and 7th term is 9
Step 1: Write the two given term equations.
a+2d=5 \quad \text{...(i)}
a+6d=9 \quad \text{...(ii)}
Step 2: Subtract equation (i) from equation (ii).
4d=4 \quad \Rightarrow \quad d=1
Step 3: Put d=1 in equation (i).
a+2(1)=5 \quad \Rightarrow \quad a=3
Final answer: The A.P. is 3, 4, 5, 6, \ldots.
Question 16: Find the 31st term if the 10th term is 38 and the 16th term is 74
Step 1: Write the equations using t_n=a+(n-1)d.
a+9d=38 \quad \text{...(i)}
a+15d=74 \quad \text{...(ii)}
Step 2: Subtract equation (i) from equation (ii).
6d=36 \quad \Rightarrow \quad d=6
Step 3: Put d=6 in equation (i).
a+9(6)=38 \quad \Rightarrow \quad a+54=38 \quad \Rightarrow \quad a=-16
Step 4: Find t_{31}.
t_{31}=a+30d=-16+30(6)=-16+180=164
Final answer: The 31st term is 164.
Examiner’s mindset for A.P. questions
In board-style Arithmetic Progression answers, the method matters as much as the number. A clear solution normally shows a, d, the formula used, substitution, simplification, and the final conclusion. For a membership question such as “Is 402 a term?”, do not stop at n=\frac{399}{5}. Add the reason: the term number must be a positive integer, so the number is not a term.
For a question with two given terms, write two equations first. Students often lose clarity by trying to guess the sequence directly. Equations such as a+9d=38 and a+15d=74 make the solution checkable.
Common mistakes in Arithmetic Progression
- Using a+nd instead of a+(n-1)d: The first term is already counted, so the common difference is added n-1 times.
- Missing the sign of d: In 12, 10, 8, 6, \ldots, the common difference is -2, not 2.
- Accepting a fractional term number: If solving l=a+(n-1)d gives n=\frac{399}{5}, that number cannot be a term position.
- Mixing the parameter with the term number: In \frac{1}{n}, \frac{n+1}{n}, \frac{2n+1}{n}, \ldots, use another symbol such as r for the 50th term position.
- Not converting mixed numbers: For 7\frac{3}{4}, first write \frac{31}{4}. This prevents errors while subtracting fractions.
Quick answer index
| Question | Final answer |
|---|---|
| 1(i) | A.P.; d=4 |
| 1(ii) | A.P.; d=-3 |
| 1(iii) | Not an A.P. |
| 1(iv) | Not an A.P. |
| 2 | 27 |
| 3 | 5, 7, 9, \ldots |
| 4 | -34 |
| 5 | 15 |
| 6 | 100\sqrt{3} |
| 7 | \frac{1}{n}+49 |
| 8 | 402 is not a term |
| 9 | d=\frac{7}{4}, t_{99}=179\frac{1}{4} |
| 10(i) | 49 terms |
| 10(ii) | 21 terms |
| 10(iii) | 10 terms |
| 11 | 52 is the 18th term |
| 12 | t_{11}=0 |
| 13 | a=13,\ d=-1 |
| 14 | 209 |
| 15 | 3, 4, 5, 6, \ldots |
| 16 | 164 |
Related ICSE Class 10 Maths study links
For more chapter practice, use the ICSE Class 10 solutions directory. To revise other Maths topics, open the ICSE Class 10 Maths study page. For exam-style practice, use the ICSE Class 10 question papers page. For board alignment, check the official CISCE website.
Frequently Asked Questions
What is the main formula in ICSE Class 10 Maths Arithmetic Progression?
The main formula in ICSE Class 10 Maths Arithmetic Progression is t_n=a+(n-1)d. Here a is the first term, d is the common difference, and t_n is the nth term.
How do I check whether a sequence is an A.P.?
Subtract each term from the next term. If the consecutive differences are equal, the sequence is an A.P. For example, 6-2=4, 10-6=4, and 14-10=4, so 2,6,10,14,\ldots is an A.P.
Why is the formula a+(n-1)d and not a+nd?
The first term a is already the first position. To reach the second term, add d once; to reach the third term, add d twice. Therefore, to reach the nth term, add d exactly n-1 times.
How are Concise Mathematics Selina Solutions Class 10 ICSE Chapter 10 Arithmetic Progression useful for revision?
Concise Mathematics Selina Solutions Class 10 ICSE Chapter 10 Arithmetic Progression are useful when you compare your working with the correct formula steps. Do not only copy the final answer; check whether your a, d, substitution, and final conclusion match the method.
What should I do if an A.P. membership question gives a fractional value of n?
If n is fractional, the given number is not a term of the A.P. A term position must be a positive integer, so a result such as n=\frac{399}{5} means the number does not occur in the sequence.