Arithmetic Progression Class 10 ICSE — Bansal Chapter 10

Written by Greya LakshmiPublished on 28 September 2026

The arithmetic progression class 10 ICSE chapter in Bansal Mathematics, Chapter 10, teaches how sequences behave when they follow a fixed rule. It builds the idea step by step: what a sequence is, how an A.P. is defined, how to find any term, how to add terms quickly, and how real-life problems become A.P.s. The chapter also covers arithmetic mean and the properties that make an A.P. easy to recognise.

Download Arithmetic Progression Class 10 ICSE Mathematics Chapter 10 PDF

Download the Arithmetic Progression Class 10 ICSE Mathematics chapter PDF from the Bansal textbook to revise offline or print for practice. It is the full chapter as printed in the book — 19 pages, 8 sections, 87 questions.

What is inside Chapter 10

19 pages · 8 sections · 23 worked examples · 87 exercise questions

Section	Title
10.1	Introduction
10.2	Arithmetic Progression (A.P.)
10.3	General Term of An Arithmetic Progression
10.4	Sum of n terms of an A.P.
10.5	Three or more terms in A.P.
10.6	Arithmetic mean
10.7	Properties of an A.P.
10.8	Word problems

EXERCISE 10(A): 17 questions
EXERCISE 10(B): 16 questions
EXERCISE 10(C): 14 questions
EXERCISE 10(D): 14 questions
EXERCISE 10(F): 6 questions
EXERCISE 10(G): 20 questions

10.1 Introduction

A sequence is a group of numbers arranged in a definite order following a rule. For example, 2, 4, 6, 8 is a sequence because each number is 2 more than the one before it. So is 30, 27, 24, where each term is 3 less than its predecessor. A list such as 5, 8, 6, 15 is not a sequence, since no rule connects the numbers.

The numbers inside a sequence are its elements or terms. When terms are joined by plus or minus signs, the sequence becomes a series, like $2 + 4 + 6 + 8$ . When the elements follow a definite rule, the series is a progression, and the pattern lets you predict the next term.

Test this idea with fresh numbers. The list 2, 5, 8, 11 is a sequence because each term is 3 more than the previous one. The list 3, 6, 12, 7 is not — 12 is double of 6, but 7 follows no such rule.

10.2 Arithmetic Progression (A.P.)

An arithmetic progression is a sequence in which each term is obtained by adding a fixed quantity to its preceding term. That fixed quantity is the common difference, written as $d$ . If the terms are $t_1, t_2, t_3, \ldots$ , then the first term is $a = t_1$ and the common difference is $d = t_2 - t_1 = t_3 - t_2 = \ldots$ .

Think of $a$ as the starting point and $d$ as the step size. The step can be positive, negative or zero. A quick mnemonic: the common difference is the next term minus the previous term. If that subtraction gives the same answer every time, the sequence is an A.P.

Check whether 14, 10, 6, 2 is an A.P. Here $10 - 14 = -4$ , $6 - 10 = -4$ and $2 - 6 = -4$ . The differences match, so it is an A.P. with $a = 14$ and $d = -4$ . The next two terms are $2 - 4 = -2$ and $-2 - 4 = -6$ .

The reverse direction works too. Given $a = 8$ and $d = 3$ , the A.P. is 8, 11, 14, 17. This introduction to sequences and series stays central to the Class 10 ICSE Mathematics book list, where later chapters build on the same notation.

10.3 General Term of An Arithmetic Progression

Write the terms of an A.P. as $a, a + d, a + 2d, a + 3d, \ldots$ . The coefficient of $d$ is always one less than the position of the term. That pattern leads to the general term:

t_n = a + (n - 1)d

Putting $n = 1, 2, 3$ into this formula gives the first, second and third terms. If the A.P. has $n$ terms altogether, its last term is $l = a + (n - 1)d$ . The sign of $d$ controls the behaviour of the A.P. If $d \gt 0$ , it is increasing. If $d \lt 0$ , it is decreasing. If $d = 0$ , all terms are the same.

Note that the textbook has a slip where it says $d = 1$ for a constant A.P.; the correct condition is $d = 0$ .

The chapter also gives a neat way to count from the end. The $r$ -th term from the end equals the $(n - r + 1)$ -th term from the beginning.

Here is an original worked pattern. For the A.P. 6, 10, 14,…, the 15th term is:

t_{15} = 6 + (15 - 1) \times 4 = 6 + 56 = 62

If the last term is given instead, use $l = a + (n - 1)d$ to solve for $n$ .

10.4 Sum of n terms of an A.P.

The sum of an A.P. comes from a pairing trick. Write the sum forwards, then write it backwards, and add the two rows. Each pair adds up to $a + l$ , and there are $n$ pairs, so $2S = n(a + l)$ . This gives the first sum formula:

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

Substitute $l = a + (n - 1)d$ to get the second form, which uses $a$ and $d$ directly:

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

Formula	Use it when you know
$S = \dfrac{n}{2}(a + l)$	first term $a$ , number of terms $n$ , last term $l$
$S = \dfrac{n}{2}[2a + (n - 1)d]$	first term $a$ , number of terms $n$ , common difference $d$

There is also a link between sums and terms: $t_n = S_n - S_{n-1}$ . The 10th term, for instance, equals the sum of 10 terms minus the sum of 9 terms.

Worked pattern: find the sum of the first 25 terms of the A.P. 3, 7, 11,… Here $a = 3$ , $d = 4$ and $n = 25$ .

$S = \frac{25}{2}[katex display="true"]2 \times 3 + 24 \times 4$ = \frac{25}{2}[6 + 96] = 25 \times 51 = 1275[/katex]

Three mistakes appear often in the Class 10 ICSE books practice work:

sign errors when $d$ is negative — $(n - 1)d$ is then subtracted, not added;
counting terms wrongly, especially when the last term is given;
picking the wrong sum formula because $l$ and $d$ are confused.

You can confirm the syllabus structure for this chapter on the official CISCE website.

10.5 Three or more terms in A.P.

When the sum of consecutive terms is given, the chapter recommends special forms that make the algebra shorter. The middle term is always $a$ , and the $d$ terms cancel out in the sum.

Three terms: $(a - d), a, (a + d)$ — sum is $3a$ ;
Four terms: $a - 3d, a - d, a + d, a + 3d$ — sum is $4a$ ;
Five terms: $(a - 2d), (a - d), a, (a + d), (a + 2d)$ — sum is $5a$ ;
Six terms: $a - 5d, a - 3d, a - d, a + d, a + 3d, a + 5d$ — sum is $6a$ .

Worked pattern: three numbers in A.P. have sum 21 and product 315. Taking them as $a - d, a, a + d$ :

3a = 21 \Rightarrow a = 7

7(49 - d^2) = 315 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2

The terms are 5, 7, 9. Reversing the order, 9, 7, 5, is equally valid.

10.6 Arithmetic mean

If three numbers $a$ , $A$ and $b$ are in A.P., then $A$ is the arithmetic mean between $a$ and $b$ . Since the gaps are equal, $A - a = b - A$ . Solving gives:

A = \frac{a + b}{2}

So the arithmetic mean of two numbers is simply their average. To insert $k$ means between $a$ and $b$ , treat the full set as an A.P. with $k + 2$ terms. The last term is $b$ , so use $l = a + (n - 1)d$ to find $d$ , then list the means one step at a time.

Worked pattern: insert three arithmetic means between 4 and 16. The set 4,…, 16 has 5 terms, so:

16 = 4 + 4d \Rightarrow d = 3

The A.P. is 4, 7, 10, 13, 16, so the three means are 7, 10 and 13.

10.7 Properties of an A.P.

Property 1: adding or subtracting the same fixed non-zero number to every term of an A.P. gives another A.P. The common difference stays unchanged.

Example: from the A.P. 6, 9, 12, 15, add 4 to each term. The result, 10, 13, 16, 19, is still an A.P. with common difference 3.

Property 2: multiplying or dividing every term by a fixed non-zero number also gives an A.P. Here the common difference changes by that same factor.

Example: multiply each term of 5, 8, 11 by 3. The result, 15, 24, 33, is an A.P. with common difference 9 — three times the original 3. This changed common difference is a common point of confusion.

Exercises extend these properties. For instance, if $a, b, c$ are in A.P., then $(b + c), (c + a), (a + b)$ are also in A.P. The pattern of this ICSE chapter carries these property questions regularly into practice work.

10.8 Word problems

Word problems follow one method: identify the sequence, confirm it is an A.P., read off $a$ , $d$ and $l$ , then apply $t_n$ or the sum formula.

The chapter applies this to simple interest. An investment of ₹8,000 at 10% per annum earns ₹800 interest in the first year, ₹1,600 in the second, ₹2,400 in the third. These amounts form an A.P. with $a = 800$ and $d = 800$ .

Rows of students work the same way. With 30 students in the first row, 27 in the second and 6 in the last, the sequence 30, 27, 24,…, 6 has $a = 30$ , $d = -3$ and $l = 6$ . Solving $30 + (n - 1)(-3) = 6$ gives $n = 9$ rows, and the sum formula gives 162 students.

Instalment problems translate each payment into one term of an A.P. Here is an original pattern: four prizes in a fund total ₹560, and each prize is ₹20 less than the previous one. The prizes form an A.P. with $n = 4$ , $d = -20$ and sum 560.

560 = \frac{4}{2}[2a + 3(-20)] \Rightarrow a = 170

The prizes are ₹170, ₹150, ₹130 and ₹110. Every term, whether a payment, a production figure or a row count, fits the same two formulas.

What to practise from Arithmetic Progression Class 10 ICSE

The practice load for this chapter is spread across six exercises. Note that the book has no Exercise 10(E) — the numbering jumps from 10(D) to 10(F).

EXERCISE 10(A) — 17 questions on identifying A.P.s and the general term;
EXERCISE 10(B) — 16 questions on term-based algebra;
EXERCISE 10(C) — 14 questions on sums of terms;
EXERCISE 10(D) — 14 questions on multiple terms and arithmetic means;
EXERCISE 10(F) — 6 word problems;
EXERCISE 10(G) — 20 mixed revision questions.

Arithmetic Progression Class 10 ICSE frequently asked questions

How do I find the common difference in an arithmetic progression class 10 ICSE problem?

Subtract any term from the term that follows it. If the terms are $t_1, t_2, t_3, \ldots$ , then $d = t_2 - t_1 = t_3 - t_2$ . The common difference can be positive, negative or zero.

When should I use S = n/2(a + l) instead of S = n/2[2a + (n − 1)d]?

Use $S = \frac{n}{2}(a + l)$ when you know the first term, the number of terms and the last term. Use $S = \frac{n}{2}[2a + (n - 1)d]$ when you know the common difference instead of the last term.

How do I find the nth term from the end of an A.P.?

The $r$ -th term from the end equals the $(n - r + 1)$ -th term from the beginning. Equivalently, rewrite the A.P. in reverse and use $t_n = a + (n - 1)d$ on the reversed sequence.

What happens if the common difference of an A.P. is negative or zero?

If $d \lt 0$ , the A.P. decreases, like 18, 16, 14. If $d = 0$ , every term is the same, so the A.P. is constant.

How are instalment and interest word problems set up as an arithmetic progression?

Each year or each month becomes one term. The yearly interest or the monthly payment changes by a fixed amount, so list the first few values, identify $a$ and $d$ , and apply the general term or sum formula.

Reference: ICSE Class 10 Mathematics textbook material, chapter 10: Arithmetic Progression — CISCE-aligned.

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