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Geometric Progression - Class 10 MCQ


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20 Questions MCQ Test - Geometric Progression

Geometric Progression for Class 10 2025 is part of Class 10 preparation. The Geometric Progression questions and answers have been prepared according to the Class 10 exam syllabus.The Geometric Progression MCQs are made for Class 10 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Geometric Progression below.
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Geometric Progression - Question 1
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Which of the following is a Geometric Progression (G.P.)?

Detailed Solution for Geometric Progression - Question 1

A G.P. is a sequence where each term is obtained by multiplying the previous term by a constant (common ratio).

For 2, 6, 18, 54 →

r = 6 ÷ 2 = 3

r = 18 ÷ 6 = 3

r = 54 ÷ 18 = 3

Ratio is constant → It is a G.P.

Geometric Progression - Question 2
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Find the common ratio of the G.P. 81, 27, 9, 3, …

Detailed Solution for Geometric Progression - Question 2

Common ratio r = any term ÷ its previous term.

r = 27 ÷ 81 = 1/3.

Checking further → 9 ÷ 27 = 1/3 and 3 ÷ 9 = 1/3.

So, r = 1/3.

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Geometric Progression - Question 3
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If the first term a = 5 and common ratio r = 2, find the 6th term.

Detailed Solution for Geometric Progression - Question 3

Formula for nth term: Tₙ = a × rⁿ⁻¹.

Here, a = 5, r = 2, n = 6.

T₆ = 5 × 2⁶⁻¹ = 5 × 2⁵.

T₆ = 5 × 32 = 160.

Geometric Progression - Question 4
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Find the 8th term of the G.P. 3, 6, 12, 24, …
Detailed Solution for Geometric Progression - Question 4

a = 3, r = 6 ÷ 3 = 2, n = 8.

T₈ = a × rⁿ⁻¹ = 3 × 2⁷.

T₈ = 3 × 128 = 384.

Geometric Progression - Question 5
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If the first term a = 729 and common ratio r = 1/3, find the 5th term.
Detailed Solution for Geometric Progression - Question 5

a = 729, r = 1/3, n = 5.

T₅ = a × rⁿ⁻¹ = 729 × (1/3)⁴.

T₅ = 729 × 1/81 = 9.

Geometric Progression - Question 6
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The 10th term of a G.P. is 512 and the first term is 1. Find the common ratio.
Detailed Solution for Geometric Progression - Question 6

T₁₀ = a × rⁿ⁻¹.
512 = 1 × r⁹.
r⁹ = 512.
Since 512 = 2⁹, r = 2.

Geometric Progression - Question 7
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Find the sum of first 8 terms of the G.P. 3, 6, 12, 24, …
Detailed Solution for Geometric Progression - Question 7

Formula: Sₙ = a × (rⁿ - 1) ÷ (r - 1), for r > 1.
a = 3, r = 2, n = 8.
S₈ = 3 × (2⁸ - 1) ÷ (2 - 1).
S₈ = 3 × (256 - 1)/1 = 3 × 255 = 765.

Geometric Progression - Question 8
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Find the sum of the first 10 terms of the G.P. 81, 27, 9, …
Detailed Solution for Geometric Progression - Question 8

a = 81, r = 1/3, n = 10.
Formula: Sₙ = a × (1 - rⁿ) ÷ (1 - r), since |r| < 1.
S₁₀ = 81 × (1 - (1/3)¹⁰) ÷ (1 - 1/3).
S₁₀ = 81 × (1 - 1/59049) ÷ (2/3).
≈ 121.

Geometric Progression - Question 9
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The sum of infinite terms of the G.P. 1, 1/2, 1/4, 1/8, … is:
Detailed Solution for Geometric Progression - Question 9

Formula for infinite sum: S∞ = a ÷ (1 - r), where |r| < 1.
a = 1, r = 1/2.
S∞ = 1 ÷ (1 - 1/2) = 1 ÷ (1/2).
S∞ = 2.

Geometric Progression - Question 10
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Find the sum to infinity of 1 - 1/4 + 1/16 - 1/64 + …
Detailed Solution for Geometric Progression - Question 10

a = 1, r = -1/4.
|r| < 1, so infinite sum exists.
S∞ = a ÷ (1 - r).
S∞ = 1 ÷ (1 - (-1/4)) = 1 ÷ (1 + 1/4).
S∞ = 1 ÷ (5/4) = 4/5.

Geometric Progression - Question 11
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If the 3rd term of a G.P. is 4 and the common ratio r = 2, find the first term.
Detailed Solution for Geometric Progression - Question 11

General term formula: Tₙ = a × rⁿ⁻¹.
Given T₃ = 4, r = 2. Substitute: 4 = a × 2³⁻¹ = a × 2².
So 4 = a × 4.
Therefore a = 4 ÷ 4 = 1.
First term a = 1.

Geometric Progression - Question 12
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Insert two geometric means between 2 and 16 to form a G.P.
Detailed Solution for Geometric Progression - Question 12

Let the sequence be 2, G₁, G₂, 16 (4 terms).
If common ratio is r, then 16 = 2 × r³ (because 4 terms mean r^(4-1) = r³).
Thus r³ = 16 ÷ 2 = 8.
So r = cube root of 8 = 2.
Then G₁ = 2 × r = 2 × 2 = 4, and G₂ = 4 × r = 4 × 2 = 8.
Inserted means: 4 and 8.

Geometric Progression - Question 13
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In a G.P., the 3rd term is 4. Find the product of the first five terms.
Detailed Solution for Geometric Progression - Question 13

Let terms be a, ar, ar², ar³, ar⁴. The 3rd term T₃ = ar² = 4.
Product P = a × ar × ar² × ar³ × ar⁴ = a⁵ × r^(0+1+2+3+4) = a⁵ × r¹⁰.
But ar² = 4 implies (ar²)⁵ = 4⁵ = a⁵ × r¹⁰.
Therefore P = (ar²)⁵ = 4⁵ = 1024.
Product of first five terms = 1024.

Geometric Progression - Question 14
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The second term of a G.P. is 2 and the sum to infinity is 8. Find the first term.
Detailed Solution for Geometric Progression - Question 14

Let first term be a and common ratio r (|r| < 1 since infinite sum exists). Given ar = second term = 2. —(i)
Sum to infinity S∞ = a ÷ (1 - r) = 8. —(ii)
From (i), a = 2 ÷ r. Substitute into (ii): (2 ÷ r) ÷ (1 - r) = 8.
Simplify: 2 ÷ [r(1 - r)] = 8 → 2 = 8r(1 - r).
Divide both sides by 2: 1 = 4r(1 - r) → 1 = 4r - 4r² → Rearranged: 4r² - 4r + 1 = 0.
This is (2r - 1)² = 0 → 2r - 1 = 0 → r = 1/2.
From (i): a = 2 ÷ r = 2 ÷ (1/2) = 4.
First term a = 4.

Geometric Progression - Question 15
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Insert two numbers between 3 and 81 so that all four are in G.P. What are the inserted numbers?
Detailed Solution for Geometric Progression - Question 15

Sequence: 3, G₁, G₂, 81 (4 terms). So 81 = 3 × r³.
Solve r³ = 81 ÷ 3 = 27 → r = cube root of 27 = 3.
G₁ = 3 × r = 3 × 3 = 9.
G₂ = 9 × r = 9 × 3 = 27.
Inserted numbers are 9 and 27.

Geometric Progression - Question 16
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Find the sum of the first 10 terms of the G.P. 96, −48, 24, −12, … (give exact fraction).
Detailed Solution for Geometric Progression - Question 16

First term a = 96. Common ratio r = (−48) ÷ 96 = −1/2.
For |r| < 1 and finite n, Sn = a × (1 − rⁿ) ÷ (1 − r). Here n = 10.
Compute r¹⁰ = (−1/2)¹⁰ = (1/2)¹⁰ = 1 ÷ 1024.
So Sn = 96 × (1 − 1/1024) ÷ (1 − (−1/2)) = 96 × (1023/1024) ÷ (3/2).
Simplify: 96 × (1023/1024) × (2/3) = (96 × 2 ÷ 1024) × (1023 ÷ 3)
→ (192 ÷ 1024) × (1023 ÷ 3) = (3/16) × 341 = 1023/16.
Sum = 1023/16.

Geometric Progression - Question 17
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Convert the recurring decimal 0.4373737... to a fraction.
Detailed Solution for Geometric Progression - Question 17

Write decimal as 0.4 + 0.0373737... The recurring part is 0.0373737...

Let x = 0.0373737... . Notice recurring block is "37" starting after two decimal places.

So x = 37/1000 + 37/1000 × (1/100) + 37/1000 × (1/100)2 + ... This is a G.P. with first term a = 37/1000 and r = 1/100.

Sum of infinite G.P. (|r| < 1): x = a ÷ (1 - r) = (37/1000) ÷ (1 - 1/100) = (37/1000) ÷ (99/100) = 37/990.

Total decimal = 0.4 + x = 4/10 + 37/990 = (396/990) + (37/990) = 433/990.

So 0.4373737... = 433/990.

Geometric Progression - Question 18
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If a G.P. has a = 2 and r = 3, find the 7th term.
Detailed Solution for Geometric Progression - Question 18

General term: Tₙ = a × rⁿ⁻¹. Here a = 2, r = 3, n = 7.
T₇ = 2 × 3⁶.
3² = 9, 3³ = 27, 3⁴ = 81, 3⁵ = 243, 3⁶ = 729.
So T₇ = 2 × 729 = 1458.
7th term = 1458.

Geometric Progression - Question 19
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How many terms are there in the G.P. 5, 15, 45, …, 1215 ?
Detailed Solution for Geometric Progression - Question 19

This is a G.P. with a = 5 and r = 15 ÷ 5 = 3. Let number of terms be n and last term l = 1215.

Formula for nth term: Tₙ = a × rⁿ⁻¹. So 1215 = 5 × 3ⁿ⁻¹.

Divide both sides: 3ⁿ⁻¹ = 1215 ÷ 5 = 243.

Recognize 243 = 3⁵. So 3ⁿ⁻¹ = 3⁵ → n − 1 = 5 → n = 6.

Number of terms = 6.

Geometric Progression - Question 20
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Does the infinite series 7, −14, 28, −56, … have a finite sum to infinity?
Detailed Solution for Geometric Progression - Question 20

The sequence is geometric with a = 7 and r = (−14) ÷ 7 = −2.

For a sum to infinity to exist, we require |r| < 1. Here |r| = 2 which is greater than 1.

Therefore the infinite series does not converge; its terms grow in magnitude and alternate sign.

Hence no finite sum to infinity exists for this series.

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