MathsCBSE Class 11 · Sequences and Series

Geometric Progression: nth Term, Sum of n Terms and Problem Types (Class 11)

Q: How do I know a sequence is a G.P.?

The ratio of each term to the one before it is the same for all terms.

Q: Can a G.P. have a negative ratio?

Yes. Then the terms alternate in sign.

Q: Why can a G.P. term be zero?

It cannot. A G.P. consists of non-zero terms, since division by a zero term would not make sense.

Q: Is the sum of squares and cubes of natural numbers examined?

The chapter page notes that special sums such as Σk, Σk² and Σk³ are assessed only formatively in 2026-27.

Updated 17 October 2026 · Fact-checked

A sequence is a G.P. if each term is the previous one multiplied by a constant ratio r. Its n-th term is a·r^(n−1) and the sum of n terms is a(r^n − 1)/(r − 1) for r ≠ 1 (n·a if r = 1). To find a and r from two terms, divide them to get r; for three numbers in G.P. take a/r, a, ar. For sums such as 7 + 77 + 777 + …, write each term as a multiple of 10^k − 1.

Each term is the one before times the same ratio r (here 2); this repeated multiplying gives a·r^(n−1) and the sum formula.

Illustration drawn for QuestionBuddy.

Understand Geometric Progression: nth Term, Sum and Problems

A geometric progression (G.P.) is a sequence of non-zero terms $a, a r, a r^{2}, \dots$ with first term $a$ and common ratio $r = \frac{a _{k + 1}}{a _{k}}$ . A sequence is a G.P. exactly when this ratio is constant.

Formulas.
$ $a_{n} = a r^{n - 1}, S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} = \frac{a ( 1 - r ^{n} )}{1 - r} (r \neq = 1), S_{n} = na (r = 1) .$ $
Use the first form for $∣ r ∣ > 1$ and the second for $∣ r ∣ < 1$ only for convenience; they are equal. A negative $r$ makes the signs alternate; $r$ between $0$ and $1$ makes the terms shrink.

**Finding $a$ and $r$ from two terms.** If $a_{p}$ and $a_{q}$ are given, divide: $\frac{a _{q}}{a _{p}} = r^{q - p}$ , solve for $r$ (take care with even powers: $r$ may be $\pm$ ), then find $a$ from $a_{p} = a r^{p - 1}$ . For $a_{3} = 12$ and $a_{6} = 96$ : $r^{3} = 8$ , $r = 2$ , $a = 3$ .

Three numbers in G.P. Take $\frac{a}{r}, a, a r$ ; their product is $a^{3}$ , which gives $a$ at once. For four numbers use $\frac{a}{r ^{3}}, \frac{a}{r}, a r, a r^{3}$ (common ratio $r^{2}$ ).

** $S_{n}$ and the terms.** If the sum of the first $n$ terms is known as a function $S_{n}$ , then $a_{1} = S_{1}$ and $a_{n} = S_{n} - S_{n - 1}$ for $n \geq 2$ ; check that the terms form a G.P. For $S_{n} = 3^{n} - 1$ : $a_{1} = 2$ , $a_{2} = 6$ , $a_{3} = 18$ , ratio $3$ , and $a_{n} = 2 \cdot 3^{n - 1}$ .

Counting terms. To find how many terms make a given sum, set $S_{n}$ equal to it and solve $r^{n} = constant$ ; the answer is a whole number or the question has no solution.

**Sums of the type $7 + 77 + 777 + \dots$ .** Write each term using $9$ s: $7 = \frac{7}{9} (10 - 1)$ , $77 = \frac{7}{9} (100 - 1)$ , and so on. Then
$ $S_{n} = \frac{7}{9} [(10 + 1 0^{2} + \dots + 1 0^{n}) - n] = \frac{7}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{7}{81} (1 0^{n + 1} - 9 n - 10) .$ $
Check with $n = 2$ : $\frac{7}{81} (1000 - 18 - 10) = 84 = 7 + 77$ .

Applications. Compound interest at rate $i$ gives a G.P. with $r = 1 + i$ ; chain messages and doubling populations are G.P.s.

Geometric Progression: nth Term, Sum and Problems: formulas to remember

n-th term: $a_{n} = a r^{(} n - 1)$; a is the first term, r the common ratio.
Sum of n terms: $S_{n} = a (r^{n} - 1) / (r - 1), r n o t e q u a l t o 1$; S_n = n a if r = 1.
Three terms in G.P.: $a / r, a, a r$; Product a³.
Terms from sums: $a_{1} = S_{1} an d a_{n} = S_{n} - S_{(} n - 1)$; For n at least 2.
Series 7 + 77 + 777 + …: $S_{n} = (7/81) (1 0^{(} n + 1) - 9 n - 10)$; From 7/9 times (10^k − 1).

How to solve Geometric Progression: nth Term, Sum and Problems questions

Use these steps for G.P. problems.

1Check or state that the ratio $r$ is constant.
2Write what is given as equations in $a$ and $r$ (use $a_{n} = a r^{n - 1}$ and $S_{n}$ ).
3Divide two equations to eliminate $a$ and find $r$ (watch for $\pm$ with even powers).
4Find $a$ and then the requested term or sum.
5For three numbers in G.P. use $\frac{a}{r}, a, a r$ .
6Check by substituting back into the data.

Geometric Progression: nth Term, Sum and Problems in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Compound interest as a G.P.

Suppose Rs 5000 is deposited at 10% per year compounded yearly. The amount at the end of each year is the previous amount times 1.1.

Find the amount after 5 years.

Try it first, then show the solutionSolution

The amounts $5500, 6050, \dots$ form a G.P. with $r = 1.1$ .
After 5 years: $5000 \times 1. 1^{5} = 5000 \times 1.61051$ .
$= 8052.55$ (rupees, to the nearest paisa).

Answer: About Rs 8052.55

Memory hook: Each year multiplies by the same factor: a G.P.

A chain message

Suppose you send a message to 3 friends. In every following round each new recipient forwards it to 3 people who have not seen it. This continues for 5 rounds.

How many people receive the message in total (excluding you)?

Try it first, then show the solutionSolution

Round 1: 3, round 2: 9, round 3: 27, ...: a G.P. with $a = 3, r = 3$ .
$S_{5} = \frac{3 ( 3 ^{5} - 1 )}{3 - 1} = \frac{3 \times 242}{2} = 363$ .

Answer: 363 people.

Memory hook: Each round multiplies by 3, so add a G.P.

Tricks to remember Geometric Progression: nth Term, Sum and Problems

Short memory aids. Say each one aloud once, then use it on a question the same day.

Divide to eliminate a
Dividing a_q by a_p removes a and leaves r^(q−p).
Use it when: Two terms given.
Middle term trick
Three numbers in G.P.: a/r, a, ar. Their product is a³.
Use it when: Sum and product problems.
Nines trick
7, 77, 777: write each as 7/9 × (10^k − 1).
Use it when: Repdigit series.
Difference of sums
a_n = S_n − S_(n−1), with a_1 = S_1.
Use it when: Sum given as a formula.

Geometric Progression: nth Term, Sum and Problems: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A term of a G.P. · 2 marks · Very short answerBoard level

Find the 8th term of the G.P. $2, 6, 18, \dots$

Try it first, then show the solutionSolution

$a = 2$ , $r = \frac{6}{2} = 3$ . (½ mark)
$a_{8} = 2 \cdot 3^{7}$ . (½ mark)
$= 2 \times 2187 = 4374$ . (1 mark)

Answer: 4374

Example 2: Sum with a negative ratio · 3 marks · Short answerBoard level

Find the sum of the first 6 terms of $3, - 6, 12, \dots$

Try it first, then show the solutionSolution

$a = 3$ , $r = - 2$ . (1 mark)
$S_{6} = \frac{3 [ 1 - ( - 2 ) ^{6} ]}{1 - ( - 2 )}$ . (1 mark)
$= \frac{3 ( 1 - 64 )}{3} = - 63$ . (1 mark)

Answer: $- 63$

Example 3: Two terms given · 3 marks · Short answerHarder board level

In a G.P. the 2nd term is 6 and the 5th term is 162. Find $a$ , $r$ and the sum of the first 6 terms.

Try it first, then show the solutionSolution

$a r = 6$ and $a r^{4} = 162$ . Dividing: $r^{3} = 27$ , so $r = 3$ . (1 mark)
$a = \frac{6}{3} = 2$ . (½ mark)
$S_{6} = \frac{2 ( 3 ^{6} - 1 )}{3 - 1} = 3^{6} - 1 = 728$ . (1½ marks)

Answer: $a = 2, r = 3, S_{6} = 728$

Example 4: Three numbers in G.P. · 4 marks · Short answerHarder board level

Three numbers in G.P. have sum 21 and product 216. Find them.

Try it first, then show the solutionSolution

Let the numbers be $\frac{a}{r}, a, a r$ . Product $a^{3} = 216$ , so $a = 6$ . (1 mark)
Sum: $\frac{6}{r} + 6 + 6 r = 21$ , so $6 r^{2} - 15 r + 6 = 0$ , i.e. $2 r^{2} - 5 r + 2 = 0$ . (1 mark)
$(2 r - 1) (r - 2) = 0$ , so $r = 2$ or $r = \frac{1}{2}$ . (1 mark)
The numbers are $3, 6, 12$ (or $12, 6, 3$ ). Check: sum $21$ , product $216$ . (1 mark)

Answer: 3, 6, 12 (in either order)

Example 5: The series 7 + 77 + 777 + … · 5 marks · Long answerHOTS / competency

Find the sum of the first $n$ terms of $7 + 77 + 777 + \dots$ and use it for $n = 3$ .

Try it first, then show the solutionSolution

Write $7 = \frac{7}{9} (10 - 1)$ , $77 = \frac{7}{9} (1 0^{2} - 1)$ , $777 = \frac{7}{9} (1 0^{3} - 1)$ , ... (1 mark)
$S_{n} = \frac{7}{9} [(10 + 1 0^{2} + \dots + 1 0^{n}) - n]$ . (1 mark)
The bracket is a G.P. with $a = 10$ , $r = 10$ : $\frac{10 ( 1 0 ^{n} - 1 )}{9}$ . (1 mark)
$S_{n} = \frac{7}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{7}{81} (1 0^{n + 1} - 9 n - 10)$ . (1 mark)
$n = 3$ : $\frac{7}{81} (10000 - 27 - 10) = \frac{7 \times 9963}{81} = 7 \times 123 = 861 = 7 + 77 + 777$ . (1 mark)

Answer: $S_{n} = \frac{7}{81} (1 0^{n + 1} - 9 n - 10)$ ; $S_{3} = 861$

Where students lose marks in Geometric Progression: nth Term, Sum and Problems

The slips examiners see most often, and what to write instead.

Mistake: Counting terms wrongly
The n-th term uses r^(n−1), not r^n.
Wrong working
8th term of $2, 6, 18, \dots$ $= 2 \cdot 3^{8}$ .
Right working
$2 \cdot 3^{7} = 4374$ .
Do this: Write "n − 1" in the exponent every time.
Usually costs: The answer.
Mistake: Losing the sign of a negative ratio
(−2)^6 is positive but (−2)^5 is negative.
Wrong working
$(- 2)^{6} = - 64$ .
Right working
$(- 2)^{6} = + 64$ .
Do this: Bracket the ratio before applying the power.
Usually costs: The answer.
Mistake: Dropping the negative ratio when solving r^2
Even powers allow ±.
Wrong working
$r^{2} = 4$ so $r = 2$ only.
Right working
$r = \pm 2$ (check which one fits the data).
Do this: Solve even powers with ±.
Usually costs: Half the answer.
Mistake: Applying a(r^n − 1)/(r − 1) with r = 1
The formula divides by zero.
Wrong working
$r = 1$ : use the formula.
Right working
$S_{n} = na$ .
Do this: Check r = 1 first.
Usually costs: The answer.

Exam tips

Always write $a$ , $r$ and $n$ before substituting into a formula.
Show the division of two equations to obtain $r$ ; it earns a method mark.
For three numbers in G.P. start with $\frac{a}{r}, a, a r$ .
For series like $7 + 77 + 777$ , show the nines step and the final check with $n = 1$ or $n = 2$ .
Verify the answer by listing the first few terms.

Practice questions on Geometric Progression: nth Term, Sum and Problems

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The 9th term of the G.P. $5, - 10, 20, \dots$ is:
(a) 1280
(b) $- 1280$
(c) 2560
(d) 640
Hint
$a_{9} = a r^{8}$ with $r = - 2$ .
Answer
(a). $5 \times (- 2)^{8} = 5 \times 256 = 1280$ .
Q2 · 1 mark · MCQHarder board level
In a G.P. the 3rd term is 5 and the 6th term is 40. The 9th term is:
(a) 160
(b) 320
(c) 640
(d) 80
Hint
Divide $a_{6}$ by $a_{3}$ to find $r^{3}$ .
Answer
(b). $r^{3} = 8$ , $r = 2$ ; $a_{9} = a_{6} \cdot r^{3} = 40 \times 8 = 320$ .
Q3 · 3 marks · Short answerHarder board level
How many terms of the G.P. $1, 3, 9, \dots$ are needed to give the sum 364?
Hint
Set $\frac{3 ^{n} - 1}{2} = 364$ .
Answer
$\frac{3 ^{n} - 1}{2} = 364$ , so $3^{n} = 729 = 3^{6}$ and $n = 6$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The sequence with $a_{n} = 3 \cdot 2^{n}$ is a G.P.
Reason (R): $\frac{a _{n + 1}}{a _{n}} = 2$ for every $n$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compute the ratio of consecutive terms.
Answer
(a). $\frac{3 \cdot 2 ^{n + 1}}{3 \cdot 2 ^{n}} = 2$ , a constant.
Q5 · 4 marks · Short answerHOTS / competency
The sum of the first $n$ terms of a sequence is $S_{n} = 3^{n} - 1$ . Show that the sequence is a G.P. and find its $n$ -th term.
Hint
Use $a_{1} = S_{1}$ and $a_{n} = S_{n} - S_{n - 1}$ .
Answer
$a_{1} = S_{1} = 2$ . For $n \geq 2$ , $a_{n} = (3^{n} - 1) - (3^{n - 1} - 1) = 3^{n - 1} (3 - 1) = 2 \cdot 3^{n - 1}$ , which also gives $a_{1} = 2$ . The ratio $\frac{a _{n + 1}}{a _{n}} = 3$ is constant, so it is a G.P. with $a_{n} = 2 \cdot 3^{n - 1}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the n-th term of a G.P.?Show answer
a r^(n−1).
2.What is the sum of n terms of a G.P. when r is not 1?Show answer
a(r^n − 1)/(r − 1).
3.What is the sum when r = 1?Show answer
n a.
4.How do you get the n-th term from S_n?Show answer
a_n = S_n − S_(n−1) for n at least 2, and a_1 = S_1.
5.How are three numbers in G.P. written?Show answer
a/r, a, ar.

Geometric Progression: nth Term, Sum and Problems: frequently asked questions

How do I know a sequence is a G.P.?

The ratio of each term to the one before it is the same for all terms.

Can a G.P. have a negative ratio?

Yes. Then the terms alternate in sign.

Why can a G.P. term be zero?

It cannot. A G.P. consists of non-zero terms, since division by a zero term would not make sense.

Is the sum of squares and cubes of natural numbers examined?

The chapter page notes that special sums such as Σk, Σk² and Σk³ are assessed only formatively in 2026-27.