Geometric Progression

Q: Find the 9th term of the GP 5, 10, 20, …

a9 = 5 × 28 = 5 × 256 = 1280.

Q: Which term of the GP 2, 2√(2), 4, … is 128?

r = √(2), so 2(√(2))n-1 = 128, giving (√(2))n-1 = 64 = (√(2))12. So n - 1 = 12 and 128 is the 13th term.

Q: The 4th term of a GP is 24 and the 7th term is 192. Find the 10th term.

(ar6)/(ar3) = r3 = (192)/(24) = 8, so r = 2 and a = (24)/(8) = 3. The 10th term is 3 × 29 = 1536.

Q: Find x if x - 2, x + 1, x + 7 are in GP. (Try it without looking back.)

(x + 1)2 = (x - 2)(x + 7) gives 2x + 1 = 5x - 14, so x = 5 and the terms are 3, 6, 12.

Q: Find the sum of the first 6 terms of the GP 4, 12, 36, …

S6 = (4(36 - 1))/(3 - 1) = 2 × 728 = 1456.

ICSE Class 10 · Mathematics · Chapter 11

What this chapter covers

Sequences that grow by multiplying
The general term of a geometric progression
Using the general term to solve problems
Three terms in a GP and the middle-term property
Sum of the first n terms of a geometric progression
Using the sum formula to find unknowns
Simple applications of geometric progression

Key points

A geometric progression is a sequence of non-zero numbers in which each term is the previous term multiplied by a fixed non-zero common ratio r.
The common ratio is r = (a_n+1)/(a_n); it may be greater than 1, between 0 and 1, negative, or 1.
The general term is a_n = ar^n-1; only r is raised to the power, and the power is one less than the position.
The nth term from the end of a finite GP with last term l is l((1)/(r))^n-1.
Non-zero numbers a, b, c are in GP exactly when b² = ac.
For three terms with a known product, take them as (a)/(r), a, ar.
The sum of the first n terms is S_n = (a(rⁿ - 1))/(r - 1), or (a(1 - rⁿ))/(1 - r), for r ≠ 1; when r = 1, S_n = na.
With the last term known, S_n = (lr - a)/(r - 1).
If the sum S_n is given, then a₁ = S₁ and a_n = S_n - S_n-1.
Growth or decrease by a fixed percentage, doubling, halving and compound interest give a GP; a fixed amount added each time gives an AP.

Questions answered

Define a geometric progression and its common ratio.

A geometric progression is a sequence of non-zero numbers in which each term after the first is obtained by multiplying the previous term by the same fixed non-zero number. That fixed number, found by dividing any term by the one before it, is the common ratio r.

Is 4, -8, 16, -32 a GP? If so, write its common ratio.

Yes. Every ratio is (-8)/(4) = (16)/(-8) = (-32)/(16) = -2, so r = -2.

Why can 0 never be a term of a GP?

The common ratio is found by dividing a term by the one before it. If a term were 0, the next ratio would need division by 0; and a non-zero first term multiplied by a non-zero ratio can never give 0.

Find the 9th term of the GP 5, 10, 20, …

a₉ = 5 × 2⁸ = 5 × 256 = 1280.

Which term of the GP 2, 2√(2), 4, … is 128?

r = √(2), so 2(√(2))^n-1 = 128, giving (√(2))^n-1 = 64 = (√(2))¹². So n - 1 = 12 and 128 is the 13th term.

The 4th term of a GP is 24 and the 7th term is 192. Find the 10th term.

(ar⁶)/(ar³) = r³ = (192)/(24) = 8, so r = 2 and a = (24)/(8) = 3. The 10th term is 3 × 2⁹ = 1536.

Find x if x - 2, x + 1, x + 7 are in GP. (Try it without looking back.)

(x + 1)² = (x - 2)(x + 7) gives 2x + 1 = 5x - 14, so x = 5 and the terms are 3, 6, 12.

Find the sum of the first 6 terms of the GP 4, 12, 36, …

S₆ = (4(3⁶ - 1))/(3 - 1) = 2 × 728 = 1456.

Past board exam questions

ICSE 2027 sample paper · 1 mark

If the n^th term of a Geometric Progression (G.P.) is given by 5 × 2ⁿ then the common ratio is:

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ICSE 2027 sample paper · 3 marks

A finite Geometric Proportion (G.P.) series contains 10 terms. The sum of the first three terms is 7 and that of the last three terms is 896. Find the first term and the common ratio.

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Let the terms be a, ar, ar², …, ar⁹. a + ar + ar² = 7 and ar⁷ + ar⁸ + ar⁹ = 896 Dividing: (ar⁷(1 + r + r²))/(a(1 + r + r²)) = (896)/(7) = 128 ⇒ r⁷ = 128 ⇒ r = 2 a(1 + 2 + 4) = 7 ⇒ a = 1 First term = 1, common ratio = 2.

ICSE 2026 board paper · 1 mark

Assertion (A): The 9^th term of a Geometric Progression (G.P.) 6, −12, 24, −48… is a positive term. Reason (R): The value of (-2)⁸ is always positive.

(A) is true and (R) is false.
(A) is false and (R) is true.
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both (A) and (R) are true but (R) is not the correct explanation of (A).

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Both (A) and (R) are true and (R) is the correct explanation of (A).

ICSE 2026 board paper · 3 marks

The fourth term of a Geometric Progression (G.P.) is 16 and its seventh term is 128. Find its: (a) common ratio (b) first term

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ar³ = 16 and ar⁶ = 128. (a) Dividing, r³ = (128)/(16) = 8 ⇒ r = 2. (b) a × 8 = 16 ⇒ a = 2.

ICSE 2025 sample paper · 4 marks

In a Geometric Progression (G.P.) the first term is 24 and the fifth term is 8. Find the ninth term of the G.P.

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a = 24, T₅ = ar⁴ = 8, so r⁴ = (8)/(24) = (1)/(3). T₉ = ar⁸ = 24 × (r⁴)² = 24 × ((1)/(3))² = (24)/(9) = (8)/(3).

ICSE 2025 compartment paper · 3 marks

If the 6th term of a series in Geometric Progression (G.P.) is 32 and the 9th term is 256, find the: (a) first term and the common ratio. (b) sum of its first 10 terms.

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(a) ar⁵ = 32, ar⁸ = 256. Dividing: r³ = 8 ⇒ r = 2. Then a = (32)/(2⁵) = 1. First term = 1, common ratio = 2. (b) S₁₀ = (a(r¹⁰ - 1))/(r - 1) = (1(1024 - 1))/(2 - 1) = 1023.

ICSE 2025 board paper · 3 marks

If 1701 is the nth term of the Geometric Progression (G.P.) 7, 21, 63 ……., find: (a) the value of 'n' (b) hence find the sum of the 'n' terms of the G.P.

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(a) a = 7, r = 3. 7 × 3^n-1 = 1701 ⇒ 3^n-1 = 243 = 3⁵ ⇒ n = 6. (b) S₆ = (7(3⁶ - 1))/(3 - 1) = (7 × 728)/(2) = 2548.

ICSE 2024 board paper · 4 marks

15, 30, 60, 120… are in G.P. (Geometric Progression). (a) Find the n^th term of this G.P. in terms of n. (b) How many terms of the above G.P. will give the sum 945?

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a = 15, r = (30)/(15) = 2. (a) T_n = ar^n-1 = 15 × 2^n-1. (b) S_n = (a(rⁿ - 1))/(r - 1) = 15(2ⁿ - 1) = 945 ⇒ 2ⁿ - 1 = 63 ⇒ 2ⁿ = 64 = 2⁶ ⇒ n = 6. 6 terms.