Geometric Progression Exe-11A Class-10 Concise ICSE Maths Solution

Geometric Progression Exe-11A Class-10 Concise ICSE Maths Solution Ch-11. In this article you would learn calculate general term of G. P. We Provide Step by Step Solutions / Answer of questions for Selina Concise Maths. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Geometric Progression Exe-11A Class-10 Concise ICSE Maths Solution Ch-11

Board	ICSE
Publications	Selina
Subject	Maths
Class	10th
Chapter-11	Geometric Progression
Writer	R.K. Bansal
Exe-11A	General Term of G. P.
Edition	2025-2026

How to find General Term of G. P.

Geometric Progression Exe-11A Class-10 Concise ICSE Maths Solution Ch-11

Que-1: Find, which of the following sequence form a G.P. :
(i) 8, 24, 72, 216, ……
(ii) 1/8, 1/24, 1/72, 1/216, ………
(iii) 9, 12, 16, 24,…..

Ans:
(i)
Given sequence: 8, 24, 72, 216, ………….
Now,
24/8=3,72/24=3,216/72=3
Since 24/8=72/24=216/72 = …….. = 3, the given sequence is a G.P. with common ratio 3.

(ii)
Given sequence: 1/8,1/24,1/72,1/216,…………….
Now,
1/24 / 1/8=13,172124=1/3 , 1/216 / 1/72=1/3
Since 1/24 / 1/8 = 1/72 / 1/24 = 1/216 / 1/72 = ……….=1/3, the given sequence is a G.P. with common ratio 1/3.

(iii)
Given sequence: 9, 12, 16, 24, …………….
Now,
12/9=4/3,16/12=4/3,24/16=3/2
Since 24/8=72/24≠216/72, the given sequence is not a G.P.

Que-2: Find the 9th term of the series : 1, 4, 16, 64,…….

Ans:
Given sequence: 1, 4, 16, 64, ……………
Now,
4/1=4,16/4=4,64/16=4
Since 4/1=16/4=64/16=……=4, the given sequence is a G.P. with the first term, a = 1 and common ratio, r = 4.
Now, T_n = ar^{n – 1}
⇒ T₉= 1 × 4^{9 – 1}
= 1 × 4⁸
= 4⁸
= 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4
= 65536

Que-3: Find the seventh term of the G.P. : 1 , √3, 3, 3√3…….

Ans: Given G.P. 1,√3,3,3√3,…………
Here,
First term, a = 1
Common ratio, r = √3/1=√3
Now, T_n = ar^{n – 1}
T₇ = 1×(√3)^7-1
= 1×(√3)^6
= (√3)^6
= 3×3×3×3×3×3
= 3 × 3 × 3
= 27

Que-4: Find the 8th term of the sequence : 3/4 , 1(1/2) , 3 , …………

Ans: Given sequence: 3/4,1(1/2),3,…………..
i.e. 3/4,3/2,3,…….
Now,
3/2 / 3/4=2,3 / 3/2=2,
Since 3/2 / 3/4=3 / 3/2=…….=2, the given sequence is a G.P. with first term, a = 3/4 and common ratio, r = 2.
Now, T_n = ar^{n – 1}
=> T₈ = ar⁷
= 34×2^(8-1)
= 34×2^7
= 34×2×2×2×2×2×2×2
= 96

Que-5: Find the 10th term of the G.P. : 12, 4, 1, 1(1/3), ………

Ans: Here,
First term, a = 12
Common ratio, r = 412=13
Now, T_n = ar^{n – 1}
=> T₁₀ = 12×(1/3)^9
= 12×1/19683
= 4/6561

Que-6: Find the nth term of the series : 1, 2, 4, 8 …….

Ans: 21=2,4/2=2,8/4=2
Since 2/1=4/2=8/4=…….=2, the given sequence is a G.P. with first term, a = 1 and common ratio, r = 2.
Now, T_n = ar^{n – 1}
=> T_n = 1 × 2^{n – 1}
= 2^{n – 1}

Que-7: Find the next three terms of the sequence : √5, 5, 5√5…..

Ans: 5/√5=√5,5√5/5=√5
Since 5/√5=5√5/52=……….=√5, the given sequence is a G.P. with first term, a = √5 and common ratio, r = √5
Now, T_n = ar^{n – 1}
∴ Next three term:
4^th term = √5×(√5)^3
= √5×5√5
= 25
5^th term = √5×(√5)^4
= √5×25
= 25√5
6^th term = √5×(√5)^5
= √5×25√5
= 125

Que-8: Find the sixth term of the series : 2^2, 2^3, 2^4,….

Ans: 2^3/2^2=2,2^4/2^3=2
Since 2^3/2^2=2^4/2^3=…….=2, the given sequence is a G.P. with first term, a = 2² = 4 and common ratio, r = 2.
Now, T_n = ar^{n – 1}
∴ T₆ = 4 × (2)⁵
= 4 × 32
= 128

Que-9: Find the seventh term of the G.P. : √3 + 1, 1,√3-1/2 ,…….

Ans: Here,
First term, a = √3+1
Common ratio, r = 1/√3+1
Now, T_n = ar^{n – 1}
=> T₇ = (√3+1)×(1/√3+1)^7-1
= (√3+1)×(1/√3+1)^6
= 1(√3+1)^5
= 1(√3+1)√5×(√3-1)^5/(√3-1)^5
= (√3-1)^5(√3-1)^5
= (√3-1)^5/32

Que-10: Find the G.P. whose first term is 64 and next term is 32.

Ans: First term, a = 64
Second term, t₂ = 32
ar = 32
64 × r = 32
r = 32/64=1/2
∴ Required G.P. = a, ar, ar^{n – 1}, ar^{n – 2}, ……….
= 64,32,64×(1/2)^2,64×(1/2)^3,……….
= 64, 32, 16, 8, …….

Que-11: Find the next three terms of the series: 2/27, 2/9, 2/3

Ans: 2/9 / 2/27=3,2/3 / 2/9=3
Since 2/9 / 2/27=2/3 / 2/9=……….=3, the given sequence is a G.P. with first term, a = 2/27 and common ratio, r = 3.
Now, T_n = ar^{n – 1}
∴ Next three terms:
4^th term =2/27×(3)^3
= 2/27×27
= 2
5^th term =2/27×(3)^4
= 2/27×27×3
= 6
6^th term = 2/27×(3)^5
= 2/27×27×9
= 18

Que-12: Find the next two terms of the series 2 – 6 + 18 – 54…..

Ans: G.P. is 2 – 6 + 18 – 54 +………
Now,
-6/2=-3,18/-6=-3,-54/18=-3
Since -6/2=18/-6=-54/18=-3, the given sequence is a G.P. with first term, a = 2 and common ratio, r = –3.
Now, T_n = ar^{n – 1}
∴ Next two terms:
5^th term = 2 × (–3)⁴
= 2 × 81
= 162
6^th term = 2 × (–3)⁵
= 2 × (–243)
= – 486