MathsICSE Class 10 · Arithmetic and Geometric Progression

Geometric Progression (G.P.): nth Term, Sum of n Terms and Solved Questions (ICSE Class 10)

Q: How do I tell an A.P. from a G.P.?

Check differences and ratios of consecutive terms. Equal differences mean an A.P.; equal ratios mean a G.P. For example 3, 6, 9 is an A.P. and 3, 6, 12 is a G.P.

Updated 15 October 2026 · Fact-checked

A geometric progression multiplies by a fixed common ratio r each time. Its nth term is Tn = arⁿ⁻¹, and the sum of its first n terms is Sn = a(rⁿ − 1)/(r − 1) for r > 1, or a(1 − rⁿ)/(1 − r) for r < 1. To find n, reduce the equation to a power of r.

Each arrow multiplies by r = 2, so T5 is four multiplications from a: Tn uses r^(n−1), but the sum uses r^n.

Illustration drawn for QuestionBuddy.

Understand Geometric Progression: nth Term and Sum

A geometric progression (G.P.) is a list in which each term is the previous term multiplied by the same non-zero number, the common ratio $r = \frac{T _{2}}{T _{1}} = \frac{T _{3}}{T _{2}} = \dots$ . An A.P. adds; a G.P. multiplies. The 2025 examiners reported candidates confusing the two, so check which pattern a list follows before choosing a formula.

General term. From $a$ , the $n$ th term has been multiplied by $r$ exactly $n - 1$ times:

$T_{n} = a r^{n - 1} .$

Careful with indices: $15 \times 2^{n - 1}$ is not $3 0^{n - 1}$ . Only the 2 is raised to the power. Both the 2024 and 2025 reports list this error.

**Sum of the first $n$ terms.**

$S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ for $r > 1$ , and $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ for $r < 1$ ( $r \neq = 1$ ).

Both are the same formula written to keep the numbers positive. Note the power is $r^{n}$ in the sum but $r^{n - 1}$ in the term. If $r = 1$ , every term is $a$ and $S_{n} = na$ .

**Finding $n$ .** Reduce to a single power and match it with a known power. In 2024 Q2(iii), $15 (2^{n} - 1) = 945$ gives $2^{n} - 1 = 63$ , $2^{n} = 64$ , $n = 6$ . In 2025 Q10(i), $7 \times 3^{n - 1} = 1701$ gives $3^{n - 1} = 243 = 3^{5}$ , $n = 6$ . Know small powers: $2^{1}$ to $2^{10}$ (2 to 1024) and $3^{1}$ to $3^{7}$ (3 to 2187).

Two terms given. Divide them: $\frac{T _{5}}{T _{2}} = \frac{a r ^{4}}{a r} = r^{3}$ . Dividing removes $a$ , just as subtracting removes $a$ in an A.P.

Three terms in G.P. $p$ , $q$ , $r$ are in G.P. exactly when $q^{2} = p r$ (the middle term squared equals the product of its neighbours).

Negative and fractional ratios. If $r < 0$ the signs alternate: 1, −2, 4, −8, … If $0 < r < 1$ the terms shrink, as in halving or depreciation problems.

Geometric Progression: nth Term and Sum: formulas to remember

Common ratio: $r = \frac{T _{2}}{T _{1}} = \frac{T _{n}}{T _{n - 1}}$; Later term divided by the earlier one.
nth term: $T_{n} = a r^{n - 1}$; Only $r$ is raised to the power.
Sum, r > 1: $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$; Power $n$ , not $n - 1$ .
Sum, r < 1: $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$; Same formula, kept positive.
Three terms in G.P.: $q^{2} = p r$; Middle term squared = product of the outer two.

How to solve Geometric Progression: nth Term and Sum questions

Confirm it is a G.P., find a and r, then use the term or sum formula and finish with powers.

1Check the pattern: equal ratios ( $\frac{T _{2}}{T _{1}} = \frac{T _{3}}{T _{2}}$ ) means a G.P.
2Write $a$ and $r$ ; for two given terms, divide to find a power of $r$ .
3For a term: $T_{n} = a r^{n - 1}$ ; work out the power first, then multiply by $a$ .
4For a sum: choose the form with a positive denominator and write $r^{n}$ in full.
5To find $n$ : divide by $a$ , isolate the power, and match it with a known power of $r$ .
6Check by listing a few terms when $n$ is small.

Geometric Progression: nth Term and Sum 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.

Growth by a constant ratio

Suppose a pond has 2000 fish and the population grows by 50% each year.

Find the population after 3 years.

Try it first, then show the solutionSolution

Growth of 50% means multiplying by 1.5 each year: $a = 2000$ , $r = \frac{3}{2}$ .
After 3 years the population is the 4th term: $2000 \times (\frac{3}{2})^{3} = 2000 \times \frac{27}{8} = 6750$ .

Answer: 6750 fish.

Memory hook: Percentage growth each period = multiply by the same ratio = G.P.

A G.P. with ratio less than 1

Imagine a medicine whose amount in the body halves every 6 hours, starting from 400 mg.

How much is left after 24 hours?

Try it first, then show the solutionSolution

24 hours is 4 halvings: amounts 400, 200, 100, 50, 25 (5 terms).
$T_{5} = 400 \times (\frac{1}{2})^{4} = 25$ .

Answer: 25 mg.

Memory hook: Count the multiplications, not the readings: 4 halvings means the 5th term.

Tricks to remember Geometric Progression: nth Term and Sum

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

A.P. adds, G.P. multiplies
A for Add, G for 'Grows by multiplying'.
Use it when: Choosing the right formula, the confusion reported in 2024 and 2025.
Only r takes the power
In $a r^{n - 1}$ , $a$ stands outside the power like a coefficient: $15 \times 2^{5} = 480$ , not $3 0^{5}$ .
Use it when: Avoiding the index error in both reports.
Term n − 1, sum n
The term formula has $r^{n - 1}$ ; the sum formula has $r^{n}$ .
Use it when: Writing the right power in each formula.

Geometric Progression: nth Term and Sum: solved examples

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

Example 1: Common ratio with fractions · 1 mark · MCQBoard level

The common ratio of the G.P. $\frac{3}{4}, \frac{3}{2}, 3, \dots$ is:
(a) $\frac{3}{4}$
(b) $\frac{1}{2}$
(c) 2
(d) $\frac{3}{2}$

Try it first, then show the solutionSolution

$r = \frac{3/2}{3/4} = \frac{3}{2} \times \frac{4}{3} = 2$ ; check $\frac{3}{3/2} = 2$ .

Answer: (c) 2

Example 2: A term and a sum · 3 marks · Short answerBoard level

For the G.P. 5, 10, 20, …, find (a) the 8th term (b) the sum of the first 8 terms.

Try it first, then show the solutionSolution

$a = 5$ , $r = 2$ .
(a) $T_{8} = 5 \times 2^{7} = 5 \times 128 = 640$ .
(b) $S_{8} = \frac{5 ( 2 ^{8} - 1 )}{2 - 1} = 5 \times 255 = 1275$ .

Answer: (a) 640 (b) 1275

Example 3: Which term, then the sum · 3 marks · Short answerHarder board level

Which term of the G.P. 2, 6, 18, … is 1458? Find the sum of the G.P. up to that term.

Try it first, then show the solutionSolution

$a = 2$ , $r = 3$ : $2 \times 3^{n - 1} = 1458 \Rightarrow 3^{n - 1} = 729 = 3^{6}$ .
$n - 1 = 6$ , so $n = 7$ .
$S_{7} = \frac{2 ( 3 ^{7} - 1 )}{3 - 1} = 3^{7} - 1 = 2187 - 1 = 2186$ .

Answer: The 7th term; sum 2186.

Example 4: G.P. from two terms · 4 marks · Short answerHarder board level

The 2nd term of a G.P. is 12 and its 5th term is 324. Find the first term, the common ratio and the sum of the first 6 terms.

Try it first, then show the solutionSolution

$a r = 12$ and $a r^{4} = 324$ . Dividing: $r^{3} = 27$ , so $r = 3$ .
$a = \frac{12}{3} = 4$ .
$S_{6} = \frac{4 ( 3 ^{6} - 1 )}{2} = 2 (729 - 1) = 1456$ .

Answer: $a = 4$ , $r = 3$ , $S_{6} = 1456$ .

Example 5: How many terms for a given sum? · 3 marks · Short answerHOTS / competency

How many terms of the G.P. 3, 6, 12, … must be added to give 765?

Try it first, then show the solutionSolution

$\frac{3 ( 2 ^{n} - 1 )}{2 - 1} = 765 \Rightarrow 2^{n} - 1 = 255$ .
$2^{n} = 256 = 2^{8}$ , so $n = 8$ .
(A common slip is to set the $n$ th term equal to 765; that is a term, not a sum.)

Answer: 8 terms.

Example 6: Case study: a doubling savings jar · 4 marks · Case-basedHOTS / competency

Meena puts ₹10 in a jar on day 1, ₹20 on day 2, ₹40 on day 3, and so on, doubling each day.
(a) How much does she put in on day 8?
(b) What is the total after 8 days?
(c) On which day does the total first exceed ₹10,000?

Try it first, then show the solutionSolution

(a) $T_{8} = 10 \times 2^{7} = 1280$ .
(b) $S_{8} = 10 (2^{8} - 1) = 2550$ .
(c) $10 (2^{n} - 1) > 10000 \Rightarrow 2^{n} > 1001$ . $2^{9} = 512$ is too small; $2^{10} = 1024$ works.
Check: $S_{9} = 5110$ , $S_{10} = 10230$ .

Answer: (a) ₹1,280 (b) ₹2,550 (c) Day 10

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

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

Mistake: Combining a with the base of the power
Only $r$ is raised to the power in $a r^{n - 1}$ .
Wrong working
$T_{n} = 15 \times 2^{n - 1} = 3 0^{n - 1}$
Right working
$T_{n} = 15 \times 2^{n - 1}$ (leave it like this)
Do this: Never multiply $a$ into the base.
Usually costs: Reported in 2024 Q2(iii) and 2025 Q10(i).
Mistake: Using the A.P. formulas for a G.P.
A G.P. multiplies; its formulas involve powers of $r$ .
Wrong working
For 15, 30, 60, …: $S_{n} = \frac{n}{2} [30 + 15 (n - 1)]$ .
Right working
$S_{n} = \frac{15 ( 2 ^{n} - 1 )}{2 - 1}$ .
Do this: Test ratios or differences before choosing a formula.
Usually costs: Reported in both years; almost all the marks.
Mistake: Wrong powers or a wrong r
Mental arithmetic with powers is error-prone, and $r$ is later ÷ earlier.
Wrong working
$3^{6} = 243$ ; $r = \frac{7}{21} = \frac{1}{3}$ .
Right working
$3^{5} = 243$ , $3^{6} = 729$ ; $r = \frac{21}{7} = 3$ .
Do this: Write out a table of powers in rough work.
Usually costs: Reported in 2025 Q10(i).

Exam tips

2024 Q2(iii) (4 marks): $T_{n} = 15 \times 2^{n - 1}$ and $15 (2^{n} - 1) = 945$ , giving $2^{n} = 64$ , $n = 6$ . Notice $2^{n} - 1$ , not $2^{n - 1}$ , in the sum.
2025 Q10(i) (3 marks): $7 \times 3^{n - 1} = 1701$ gives $n = 6$ , then $S_{6} = 2548$ .
When finding $n$ , write the power as an exact power (for example $64 = 2^{6}$ ) and state $n$ .
Use the $r > 1$ form of the sum for growing G.P.s so the numerator and denominator are positive.

Practice questions on Geometric Progression: nth Term and Sum

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 6th term of the G.P. 1, −2, 4, … is:
(a) 32
(b) −32
(c) 64
(d) −64
Hint
$r = - 2$ ; $T_{6} = (- 2)^{5}$ .
Answer
(b) −32
Q2 · 2 marks · Short answerBoard level
Find the sum of the first 5 terms of the G.P. 4, 12, 36, …
Hint
$a = 4$ , $r = 3$ .
Answer
$\frac{4 ( 3 ^{5} - 1 )}{2} = 2 \times 242 = 484$
Q3 · 3 marks · Short answerHarder board level
Find the sum $1 + \frac{1}{3} + \frac{1}{9} + \dots$ to 5 terms.
Hint
Use $\frac{a ( 1 - r ^{n} )}{1 - r}$ with $r = \frac{1}{3}$ .
Answer
$\frac{1 - \frac{1}{243}}{\frac{2}{3}} = \frac{242}{243} \times \frac{3}{2} = \frac{121}{81}$
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): 2, 6, 12, 20 is a G.P.
Reason (R): In a G.P., each term is obtained by multiplying the previous term by the same number.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Compare $\frac{6}{2}$ and $\frac{12}{6}$ .
Answer
(b). The ratios are 3 and 2, so it is not a G.P.; (R) is the correct definition.
Q5 · 3 marks · Short answerHarder board level
Find $x$ if $x - 1$ , $x + 1$ and $x + 7$ are consecutive terms of a G.P.
Hint
$(x + 1)^{2} = (x - 1) (x + 7)$ .
Answer
$x^{2} + 2 x + 1 = x^{2} + 6 x - 7$ gives $x = 2$ ; the terms are 1, 3, 9.
Q6 · 4 marks · Short answerHOTS / competency
The first term of a G.P. is 1 and the sum of its first three terms is 13. Find the possible common ratios and write each G.P.
Hint
$1 + r + r^{2} = 13$ .
Answer
$r^{2} + r - 12 = 0$ , so $r = 3$ or $r = - 4$ : 1, 3, 9, … or 1, −4, 16, …
Q7 · 3 marks · Short answerHarder board level
How many terms of the G.P. 2, 4, 8, … add up to 1022?
Hint
$2 (2^{n} - 1) = 1022$ .
Answer
$2^{n} - 1 = 511$ , $2^{n} = 512$ , $n = 9$ .

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.Write the nth term of a G.P.Show answer
$T_{n} = a r^{n - 1}$
2.Find the 5th term of 3, 6, 12, …Show answer
$3 \times 2^{4} = 48$
3.What is $r$ for 8, 4, 2, …?Show answer
$\frac{1}{2}$
4.When are $p$ , $q$ , $r$ in G.P.?Show answer
When $q^{2} = p r$ .
5.Write $S_{n}$ for a G.P. with $r > 1$ .Show answer
$\frac{a ( r ^{n} - 1 )}{r - 1}$

Geometric Progression: nth Term and Sum: frequently asked questions

What is the nth term of a geometric progression?

Tn = arⁿ⁻¹, where a is the first term and r is the common ratio. For example, the 8th term of 5, 10, 20, … is 5 × 2⁷ = 640.

What is the sum of n terms of a G.P.?

Sn = a(rⁿ − 1)/(r − 1) when r > 1, and a(1 − rⁿ)/(1 − r) when r < 1. If r = 1, every term equals a and Sn = na.

How do I tell an A.P. from a G.P.?

Check differences and ratios of consecutive terms. Equal differences mean an A.P.; equal ratios mean a G.P. For example 3, 6, 9 is an A.P. and 3, 6, 12 is a G.P.