Geometric Progression Class 10 ICSE — Bansal Chapter 11

Written by Greya LakshmiPublished on 28 September 2026

Geometric progression class 10 ICSE from Bansal Mathematics explains sequences where each term comes from multiplying the previous term by a fixed quantity. This chapter covers the nth term formula, the sum of n terms, infinite sums, recurring decimals, and geometric means. You will learn each formula and how to apply it to typical exam problems step by step. The section-by-section walkthrough below follows the book’s own order.

Download Geometric Progression Class 10 ICSE Mathematics Chapter 11 PDF

Download the Geometric 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 — 23 pages, 9 sections, 44 questions.

What is inside Chapter 11

23 pages · 9 sections · 26 worked examples · 44 exercise questions

Section	Title
11.1	Introduction
11.2	Geometric progression
11.3	General term of a geometric progression
11.4	Properties of geometric progression
11.5	Sum of n terms of a G.P.
11.6	Sum of Infinite Terms of a G.P.
11.7	Recurring decimals
11.8	Geometric mean between numbers a and b
11.9	n Geometric means between a and b

EXERCISE 11(A): 12 questions
EXERCISE 11(B): 10 questions
EXERCISE 11(C): 8 questions
EXERCISE 11(F): 4 questions
EXERCISE 11(G): 10 questions

11.1 Introduction

A sequence is an arrangement of numbers written in a definite order according to a given rule. For example, 2, 6, 18, 54,… is a sequence where each term multiplied by 3 gives the next term. Another example is $1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \ldots$ , where each term is multiplied by $-\frac{1}{2}$ . A third example is 6, 4, 2, 0,…, where each term is decreased by 2.

When the terms of a sequence are connected by plus or minus signs, you get a series. So $2 + 6 + 18 + 54 + \ldots$ is a series, and so is $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \ldots$ .

In general, the chapter treats a sequence and a series as the same thing. The key idea is that a definite rule connects consecutive terms. Once you spot the rule, you can predict any term in the sequence.

A fresh example: the sequence 5, 25, 125,… follows the rule “multiply by 5”. Written as a series, it is $5 + 25 + 125 + \ldots$ .

11.2 Geometric progression

A geometric progression (abbreviated G.P.) is a sequence in which each term is obtained by multiplying (or dividing) its preceding term by a fixed quantity. That fixed quantity is called the common ratio, denoted by $r$ .

To find $r$ , divide any term by its preceding term. For the G.P. 16, 8, 4, 2, 1, $\frac{1}{2}$ , each ratio is $\frac{1}{2}$ :

$\frac{8}{16} = \frac{4}{8} = \frac{2}{4} = \dots = \frac{1}{2} = r$

The common ratio can be negative. For 60, $-24$ , $9\frac{3}{5}$ , $-3\frac{21}{25}$ , the ratio is $-\frac{2}{5}$ . A negative $r$ makes the terms alternate in sign, so plus and minus terms follow each other in turn.

Test a fresh sequence: 4, $-12$ , 36, $-108$ . Divide each term by its predecessor:

$\frac{-12}{4} = -3, \quad \frac{36}{-12} = -3, \quad \frac{-108}{36} = -3$

All ratios equal $-3$ , so this is a G.P. with $r = -3$ . If the first term is 5 and $r = 3$ , the G.P. is 5, 15, 45, 135,…. As a series, write $5 + 15 + 45 + 135 + \ldots$ .

11.3 General term of a geometric progression

Let the first term of a G.P. be $a$ and the common ratio be $r$ . Then the terms build up as:

$t_1 = a = ar^{1-1}, \quad t_2 = ar = ar^{2-1}, \quad t_3 = ar^2 = ar^{3-1}, \quad t_4 = ar^3 = ar^{4-1}$

The pattern generalises to the nth term:

$t_n = ar^{n-1}$

If the G.P. has exactly $n$ terms, the nth term is called the last term, denoted by $l$ :

$l = ar^{n-1}$

Any three of $a$ , $r$ , $n$ , and $t_n$ determine the fourth. So the 8th term is $ar^7$ , the 15th term is $ar^{14}$ , and the 20th term is $ar^{19}$ .

Worked pattern 1: Find the 10th term of 2, 6, 18,…. Here $a = 2$ and $r = 3$ :

$t_{10} = 2 \times 3^{9} = 2 \times 19683 = 39366$

Worked pattern 2: Which term of 6, 12, 24,… equals 768? Here $a = 6$ and $r = 2$ . Solve:

$6 \times 2^{n-1} = 768$

Divide by 6: $2^{n-1} = 128 = 2^7$ . So $n - 1 = 7$ and $n = 8$ . The 8th term is 768.

Worked pattern 3: The 3rd term is 20 and the 6th term is 160. Then $ar^2 = 20$ and $ar^5 = 160$ . Dividing gives $r^3 = 8$ , so $r = 2$ . Then $a \times 4 = 20$ , so $a = 5$ . This is the kind of problem you will meet often in the Class 10 ICSE syllabus.

11.4 Properties of geometric progression

A G.P. has eight properties that simplify proofs and speed up problem solving.

The ratio of consecutive terms is always the same: $\frac{t_2}{t_1} = \frac{t_3}{t_2} = \frac{t_4}{t_3} = \ldots$ .
The product of two terms equidistant from the beginning and the end equals the first term times the last term.
If $a, b, c$ are in G.P., then $b^2 = ac$ . Conversely, if $b^2 = ac$ , then $a, b, c$ are in G.P. For 2, 6, 18, check $6^2 = 36$ and $2 \times 18 = 36$ .
Terms at equal distances from each other stay in G.P. For 2, 6, 18, 54, 162, 486, the terms 2, 18, 162 form a G.P.
Multiplying or dividing every term by a non-zero number keeps a G.P.
The reciprocals of the terms of a G.P. also form a G.P. For 6, 24, 96, 384, the reciprocals $\frac{1}{6}, \frac{1}{24}, \frac{1}{96}, \frac{1}{384}$ form a G.P.
Raising every term to the same non-zero power keeps a G.P. For 2, 8, 32, the fifth powers $2^5, 8^5, 32^5$ form a G.P.
Multiplying (or dividing) corresponding terms of two G.P.s gives a G.P. If $a, ar, ar^2, \ldots$ and $A, AR, AR^2, \ldots$ are G.P.s, then $aA, arAR, ar^2AR^2, \ldots$ is also a G.P.

These properties link G.P. with the arithmetic progression (A.P.) studied in the previous chapter. Many proof questions combine both. For example, if $a, b, c$ are in G.P., then $\frac{1}{a+b}, \frac{1}{2b}$ , and $\frac{1}{b+c}$ turn out to be in A.P.

11.5 Sum of n terms of a G.P.

Start from the sum of $n$ terms:

$S_n = a + ar + ar^2 + ar^3 + \dots + ar^{n-1}$

The formula splits into three cases based on the value of $r$ .

Case	Formula	When to use	Common mistake
$r = 1$	$S_n = na$	All terms equal	Forgetting this case exists
$r \lt 1$	$S_n = \frac{a(1 - r^n)}{1 - r}$	Common ratio below 1	Using the $r \gt 1$ form instead
$r \gt 1$	$S_n = \frac{a(r^n - 1)}{r - 1}$	Common ratio above 1	Swapping numerator terms

There are also last-term forms. When $r \lt 1$ :

$S_n = \frac{a - lr}{1 - r}$

When $r \gt 1$ :

$S_n = \frac{lr - a}{r - 1}$

Worked pattern 1: Find the sum of 8 terms of $4 + 12 + 36 + \ldots$ . Here $a = 4$ , $r = 3$ , and $r \gt 1$ :

$S_8 = \frac{4(3^8 - 1)}{3 - 1} = \frac{4(6561 - 1)}{2} = 13120$

Worked pattern 2: The series $6 + 66 + 666 + \ldots$ to $n$ terms. Factor out 6, then write each term as a difference:

$6(1 + 11 + 111 + \ldots) = \frac{6}{9}(9 + 99 + 999 + \ldots)$

$= \frac{6}{9}\left[katex display="true"](10 - 1) + (10^2 - 1) + \ldots\right$ = \frac{6}{9}\left $\frac{10(10^n - 1)}{9} - n\right$ [/katex]

$= \frac{20}{27}(10^n - 1) - \frac{2}{3}n$

Common mistakes in this section include picking the wrong case and misreading the last term $l$ when using the last-term forms.

11.6 Sum of Infinite Terms of a G.P.

When the numerical value of $r$ is less than 1, written $|r| \lt 1$ or $-1 \lt r \lt 1$ , the sum of infinite terms of a G.P. is:

$S = \frac{a}{1 - r}$

Why does this work? As $n$ grows large, $r^n$ approaches zero when $|r| \lt 1$ . So the $r^n$ term in the finite formula vanishes, leaving $\frac{a}{1 - r}$ .

Edge case: when $|r| \geq 1$ , the series has no finite sum. The terms grow or stay constant, so the partial sums do not settle on a limit. For example, $1 + 2 + 4 + 8 + \ldots$ grows without bound. This is the edge case to remember for this ICSE Class 10 chapter.

Common mistakes:

Applying $S = \frac{a}{1 - r}$ without first checking $|r| \lt 1$ .
Confusing $r \lt 1$ with $|r| \lt 1$ when $r$ is negative. For $r = -2$ , the statement $r \lt 1$ is true, but $|r| = 2$ , so no finite sum exists.
Forgetting to handle the negative sign in the denominator when $r$ is negative.

Worked pattern 1: Find the sum of $1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots$ . Here $a = 1$ and $r = -\frac{1}{3}$ . Since $|r| = \frac{1}{3} \lt 1$ :

$S = \frac{1}{1 - (-\frac{1}{3})} = \frac{1}{1 + \frac{1}{3}} = \frac{3}{4}$

Worked pattern 2: The first term of a G.P. is 6 and its infinite sum is 9. Solve:

$9 = \frac{6}{1 - r}$

So $1 - r = \frac{2}{3}$ and $r = \frac{1}{3}$ . Check $|r| = \frac{1}{3} \lt 1$ , so the sum is valid.

11.7 Recurring decimals

To evaluate a recurring decimal, split it into a terminating part plus an infinite G.P. of the repeating block. Then apply $S = \frac{a}{1 - r}$ .

Worked pattern: Evaluate 0.327 with the bar over 27, meaning $0.3272727\ldots$ . Write it as:

$0.327 = 0.3 + 0.027 + 0.00027 + \ldots$

The repeating block 27 gives a G.P. with first term $\frac{27}{1000}$ and common ratio $\frac{1}{100}$ :

$0.327 = \frac{3}{10} + \frac{27}{1000}\left(1 + \frac{1}{100} + \frac{1}{10000} + \ldots\right)$

$= \frac{3}{10} + \frac{27}{1000} \times \frac{1}{1 - \frac{1}{100}} = \frac{3}{10} + \frac{27}{1000} \times \frac{100}{99}$

$= \frac{3}{10} + \frac{27}{990} = \frac{297 + 27}{990} = \frac{324}{990} = \frac{18}{55}$

So $0.327 = \frac{18}{55}$ . The common ratio between successive repeating blocks is always $\frac{1}{100}$ for a two-digit repeating block. This is the ICSE use of infinite G.P. sums to turn recurring decimals into exact fractions.

11.8 Geometric mean between numbers a and b

If $a$ and $b$ are two positive numbers, and $a, G, b$ are in G.P., then:

$\frac{G}{a} = \frac{b}{G}$

Cross-multiplying gives $G^2 = ab$ , so:

$G = \sqrt{ab}$

$G$ is called the geometric mean between $a$ and $b$ . The condition that $a$ and $b$ are positive matters. The square root of a negative product is not a real number, so the geometric mean is defined only for positive numbers.

Worked pattern 1: The geometric mean between 8 and 18 is:

$G = \sqrt{8 \times 18} = \sqrt{144} = 12$

Worked pattern 2: The geometric mean between 2 and 32 is:

$G = \sqrt{2 \times 32} = \sqrt{64} = 8$

Check that 2, 8, 32 form a G.P. with $r = 4$ . The middle term 8 is the geometric mean.

11.9 n Geometric means between a and b

To insert $n$ geometric means between $a$ and $b$ , treat $b$ as the $(n+2)$ th term of the G.P. So:

$b = ar^{n+1}$

Solve for $r$ , then the means are $ar, ar^2, \ldots, ar^n$ .

Worked pattern 1: Insert two numbers between 2 and 54. Here $n = 2$ , $a = 2$ , $b = 54$ :

$54 = 2r^3$

So $r^3 = 27$ and $r = 3$ . The means are $2 \times 3 = 6$ and $2 \times 9 = 18$ . The full G.P. is 2, 6, 18, 54.

When the product of three terms in G.P. is known, write them as $\frac{a}{r}, a, ar$ . The product is $a^3$ , so you find $a$ quickly.

Worked pattern 2: Three numbers in G.P. have product 27 and sum 13. Write them as $\frac{a}{r}, a, ar$ . Then $a^3 = 27$ , so $a = 3$ . The sum is:

$\frac{3}{r} + 3 + 3r = 13$

Multiply by $r$ : $3 + 3r + 3r^2 = 13r$ , so $3r^2 - 10r + 3 = 0$ . This gives $r = 3$ or $r = \frac{1}{3}$ . Either way, the terms are 1, 3, 9.

What to practise from Geometric Progression Class 10 ICSE

The chapter has five exercise blocks. Each one drills a specific skill, so plan your practice around what you find hardest.

Exercise	Questions	Skill drilled
EXERCISE 11(A)	12	Identifying G.P.s and finding the nth term
EXERCISE 11(B)	10	Finding terms and the G.P. from given conditions
EXERCISE 11(C)	8	Terms from the end and proof questions
EXERCISE 11(F)	4	Infinite sums and recurring decimals
EXERCISE 11(G)	10	Geometric means and inserting means

For the full set of chapters in this subject, visit the Class 10 Mathematics books hub.

Geometric Progression Class 10 ICSE frequently asked questions

What is a geometric progression class 10 icse?

A geometric progression is a sequence where each term is obtained by multiplying the preceding term by a fixed quantity called the common ratio. For example, 2, 6, 18, 54 is a G.P. with common ratio 3.

What is the nth term formula for a geometric progression?

The nth term of a G.P. with first term $a$ and common ratio $r$ is $t_n = ar^{n-1}$ . The last term of an n-term G.P. is the same expression, $l = ar^{n-1}$ .

When can you use the infinite sum formula for a geometric progression?

Use $S = \frac{a}{1 - r}$ only when the absolute value of the common ratio is less than 1, written $|r| \lt 1$ . When $|r| \geq 1$ , the infinite sum does not exist.

How do you convert a recurring decimal into a fraction using a geometric progression?

Write the decimal as a terminating part plus an infinite G.P. of the repeating block, then apply $S = \frac{a}{1 - r}$ . For 0.327 with repeating 27, the result is $\frac{18}{55}$ . You can verify the syllabus scope for this chapter on the official CISCE website.

How do you insert geometric means between two numbers in this ICSE chapter?

Treat $b$ as the $(n+2)$ th term of the G.P. and solve $b = ar^{n+1}$ for $r$ . Then the means are $ar, ar^2, \ldots, ar^n$ . For two means between 2 and 54, solve $54 = 2r^3$ , giving $r = 3$ and the means 6 and 18.

Reference: ICSE Class 10 Mathematics textbook material, chapter 11: Geometric Progression — CISCE-aligned.

Explore Class 10 Maths Notes

Related chapters: