Question 1

The $5$th term of the GP $2,\,6,\,18,\dots$ is:

Accepted Answer

**Answer: (a) $162$.** Here $a=2,\ r=3$, so $T_5=ar^{4}=2	imes3^{4}=2	imes81=162$.

Question 2

In a GP the $3$rd term is $12$ and the $6$th term is $96$. The common ratio is:

Accepted Answer

**Answer: (a) $2$.** $\dfrac{T_6}{T_3}=\dfrac{ar^{5}}{ar^{2}}=r^{3}=\dfrac{96}{12}=8$, so $r=2$.

Question 3

Which term of the GP $3,\,6,\,12,\dots$ is $384$?

Accepted Answer

**Answer: (b) $8$th.** $ar^{n-1}=384\Rightarrow 3	imes2^{\,n-1}=384\Rightarrow 2^{\,n-1}=128=2^{7}$, so $n-1=7,\ n=8$.

Question 4

The sum of the first $6$ terms of the GP $1,\,-2,\,4,\,-8,\dots$ is:

Accepted Answer

**Answer: (a) $-21$.** $a=1,\ r=-2$. $S_6=\dfrac{a(r^{6}-1)}{r-1}=\dfrac{1(64-1)}{-2-1}=\dfrac{63}{-3}=-21$.

Question 5

**Assertion (A):** The sequence $5,\,10,\,20,\,40,\dots$ is a GP with common ratio $2$. **Reason (R):** A sequence is a GP if the ratio of every term to its preceding term is constant.

Accepted Answer

**Answer: (a)** Each ratio $\dfrac{10}{5}=\dfrac{20}{10}=\dfrac{40}{20}=2$ is constant, so it is a GP with $r=2$, and R correctly explains A.

Question 6

Find the $8$th term of the GP $5,\,10,\,20,\dots$

Accepted Answer

$a=5,\ r=2$. $T_8=ar^{7}=5	imes2^{7}=5	imes128=640.$

Question 7

Find the sum: $1+3+9+\dots+2187$.

Accepted Answer

$a=1,\ r=3$. The last term $2187=3^{7}=ar^{n-1}$ gives $n=8$. $S_8=\dfrac{a(r^{n}-1)}{r-1}=\dfrac{1(3^{8}-1)}{3-1}=\dfrac{6561-1}{2}=\dfrac{6560}{2}=3280.$

Question 8

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

Accepted Answer

$T_4=ar^{3}=24$ and $T_7=ar^{6}=192$. Dividing, $r^{3}=\dfrac{192}{24}=8\Rightarrow r=2$. Then $a	imes8=24\Rightarrow a=3$. The GP is $3,\,6,\,12,\,24,\dots$

Question 9

How many terms of the GP $2,\,4,\,8,\dots$ add up to $510$?

Accepted Answer

$a=2,\ r=2$. $S_n=\dfrac{a(r^{n}-1)}{r-1}=\dfrac{2(2^{n}-1)}{2-1}=2(2^{n}-1)=510.$ $2^{n}-1=255\Rightarrow 2^{n}=256=2^{8}\Rightarrow n=8.$ So $8$ terms are needed.

Question 10

The sum of three numbers in GP is $21$ and their product is $216$. Find the numbers.

Accepted Answer

Let the numbers be $\dfrac{a}{r},\,a,\,ar$ .

Product: $\dfrac{a}{r}\times a\times ar=a^{3}=216\Rightarrow a=6.$

Sum: $6\left(\dfrac1r+1+r\right)=21\Rightarrow \dfrac1r+r=\dfrac{21}{6}-1=\dfrac{5}{2}.$

$2r^{2}-5r+2=0\Rightarrow (2r-1)(r-2)=0\Rightarrow r=2\ \text{or}\ \tfrac12.$

The numbers are $3,\,6,\,12.$

Question 11

In a GP the $2$nd term is $9$ and the $4$th term is $81$ (all terms positive). Find (i) the common ratio, (ii) the first term, and (iii) the sum of the first $6$ terms.

Accepted Answer

(i) $T_2=ar=9$ and $T_4=ar^{3}=81$. $\dfrac{ar^{3}}{ar}=r^{2}=\dfrac{81}{9}=9\Rightarrow r=3$ (positive). (ii) $ar=9\Rightarrow a	imes3=9\Rightarrow a=3.$ (iii) $S_6=\dfrac{a(r^{6}-1)}{r-1}=\dfrac{3(3^{6}-1)}{3-1}=\dfrac{3(729-1)}{2}=\dfrac{3	imes728}{2}=1092.$

Question 12

Going back through the generations, a person has $2$ parents, $4$ grandparents, $8$ great-grandparents, and so on, the number doubling each generation. Find (i) the number of ancestors in the $10$th generation back, and (ii) the total number of ancestors over these $10$ generations.

Accepted Answer

The counts form a GP $2,\,4,\,8,\dots$ with $a=2,\ r=2$ .

(i) $T_{10}=ar^{9}=2\times2^{9}=2^{10}=1024$ ancestors.

(ii) $S_{10}=\dfrac{a(r^{10}-1)}{r-1}=\dfrac{2(2^{10}-1)}{2-1}=2(1024-1)=2046$ ancestors.

Question 13

A student starts a $10$-day saving challenge. On day $1$ she saves Rs $1$, on day $2$ Rs $2$, on day $3$ Rs $4$, and the amount doubles each day. (i) How much does she save on day $5$? (ii) How much does she save on day $10$? (iii) What is her total saving over the $10$ days?

Accepted Answer

The daily amounts form a GP with $a=1,\ r=2$ .

(i) Day $5$ : $T_5=ar^{4}=1\times2^{4}=$ Rs $16.$

(ii) Day $10$ : $T_{10}=ar^{9}=2^{9}=$ Rs $512.$

(iii) Total: $S_{10}=\dfrac{a(r^{10}-1)}{r-1}=\dfrac{1(2^{10}-1)}{2-1}=1023$ , i.e. Rs $1023.$

Question 14

Are these Geometric Progression important questions free?

Accepted Answer

Yes. All 13 ICSE Class 10 Maths important questions for Geometric Progression are free, with full model answers and no login required.

Question 15

Do these Geometric Progression questions follow the latest ICSE syllabus?

Accepted Answer

Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

Question 16

How should I practise the Geometric Progression important questions?

Accepted Answer

Attempt each question on paper first, then reveal the model answer to check your method — not just the final result. Re-do anything you got wrong the same day.

Question 17

What types of questions are covered for Geometric Progression?

Accepted Answer

A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

Question type	Count	Marks
MCQ	4	1
Assertion–Reason	1	1
Very Short	2	2
Short Answer	3	3
Long Answer	2	5
Case-based	1	4

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