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

Geometric Progression Exe-11B Class-10 Concise ICSE Maths Solution Ch-11.  In this article you would learn to solve more tough problems on 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-11B 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-11B	More on General Term of G. P.
Edition	2025-2026

More on General Term of G. P.

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

Que-1: Which term of the G.P. : – 10, 5/√3, -5/6, ……is -5/72 ?

Ans: First term, a = –10
Common ratio, r = 5/√3 / -10=-1/2√3
If -5/72 is the n^th term of the given G.P., then -5/72 = ar^{n – 1}
-5/72=-10×(1/2√3)^n-1
1/144=(1/2√3)^n-1
1/2×2×2×2×√3×√3×√3×√3=(1/2√3)^n-1
n=>(1/2√3)^4=(1/2√3)^n-1
n – 1 = 4
n = 4 + 1
n = 5

Que-2: The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.

Ans: Let the first term of the G.P. be a and its common ratio be r.
5^th term = 81 ⇒ ar⁴ = 81
2^nd term = 24 ⇒ ar = 24
Now, ar^4/ar=81/24
r^3=27/8
r=3/2
ar = 24
a×3/2=24
a = 16
∴ G.P. = a, ar, ar², ar³, …………..
= 16,24,16×(32)^2,16×(3/2)^3,………
= 16, 24, 36, 54, …..

Que-3: Fourth and seventh terms of a G.P. are 1/18 and -1/486 respectively. Find the GP.

Ans: Let the first term of the G.P. be a and its common ratio be r.
4^th term =1/18=>ar^3=1/18
7^th term =148/6=>ar^6=1/486
Now, ar^6/ar^3=-1/486 / 1/18
r^3=-1/27
r=-1/3
ar^3=1/18
a×(-1/3)^3=1/18
a=-27/18=-3/2
∴ G.P. = a, ar, ar², ar³, …….
= -3/2,-3/2×(-1/3),-3/2×(-1/3)^2,1/18,…….
= -3/2, 1/2, -1/6, 1/18, ………

Que-4: If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term

Ans: Let the first term of the G.P. be a and its common ratio be r.
∴ 1^st term = a = 2
And 3^rd term = 8 => ar² = 8
Now, ar^2/a=8/2
r² = 4
r = 2
When a = 2 and r = 2
2^nd term = ar = 2 × 2 = 4

Que-5: The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.

Ans: Let the first term of the G.P. be a and its common ratio be r.
Now,
t₃ × t₈ = 243
ar² × ar⁷ = 243
a²r⁹ = 243  …(i)
Also,
t₄ = 3
ar³ = 3
a=3/r^3
Substituting the value of a in (i), we get
(3/r^3)^2×r^9=243
9/r^6×r^9=243
r³ = 27
r = 3
a=3/3^3
= 3/27
= 1/9
∴ 7^th term = t₇
= ar⁶
= 19×(3)^6
= 81

Que-6: Find the geometric progression with 4th term = 54 and 7th term = 1458.

Ans: Let the first term of the G.P. be a and its common ratio be r.
4^th term = 54 ⇒ ar³ = 54
7^th term = 1458 ⇒ ar⁶ = 1458
Now, ar^6/ar^3=1458/54
r³ = 27
r = 3
ar³ = 54
a × (3)³ = 54
a=54/27=2
∴ G.P. = a, ar, ar², ar³, ……..
= 2, 2 × 3, 2 × (3)², 54, ………..
= 2, 6, 18, 54, …….

Que-7: Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.

Ans: Let the first term of the G.P. be a and its common ratio be r.
Now, 2^nd term = t₂ = 6 => ar = 6
Also, t₅ = 9 × t₃
ar⁴ = 9 × ar²
r² = 9
r = ±3
Since, each term of a G.P. is positive, we have r = 3 and ar = 6
a × 3 = 6
a = 2
∴ G.P. = a, ar, ar², ar³, ……..
= 2, 6, 2 × (3)², 2 × (3)³, …………
= 2, 6, 18, 54, ……..

Que-8: The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.

Ans: Let the first term of the G.P. be a and its common ratio be r.
4^th term  = t₄ = 10 ⇒ ar³ = 10
7^th term = t₇ = 80 ⇒ ar⁶ = 80
ar^6/ar^3=80/10
r³ = 8
r = 2
ar³ = 10
a × (2)³ = 10
a=10/8=5/4
Last term = I = 2560
Let there be n term in given G.P.
t_n = 2560
ar^{n – 1} = 2560
5/4×(2)^n-1=2560
(2)^{n – 1} = 2048
(2)^{n – 1} = (2)¹¹
n – 1 = 11
n = 12
Thus, we have
First term = 5/4 common ratio = 2 and number of terms = 12

Que-9: If the 4th and 9th terms of a G.P. are 54 and 13122 respectively, find the GP. Also, find its general term.

Ans: Let the first term of the G.P. be a and its common ratio be r.
4^th term = t₄ = 54 ⇒ ar³ = 54
9^th term = t₉ = 13122 ⇒ ar⁸ = 13122
ar^8/ar^3=13122/54
r⁵ = 243
r = 3
ar³ = 54
a × (3)³ = 54
a=54/27=2
∴ Required G.P. = a, ar, ar², ar³, ……..
= 2, 2 × 3, 2 × (3)², 54
= 2, 6, 18, 54
General term = t_n
= ar^{n – 1}
= 2 × (3)^{n – 1}

Que-10: The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : q² = pr.

Ans: Let the first term of the G.P. be a and its common ratio be r.
5^th term = t₅ = p
=> ar⁴ = p
8^th term = t₈ = q
=> ar⁷ = q
11^th term = t₁₁ = r
=> ar¹⁰ = r
Now,
pr = ar⁴ × ar¹⁰
= a² × r¹⁴
= (a × r⁷)²
= q²
=> q² = pr

— :  Geometric Progression Exe-11B Class-10 Concise ICSE Maths step by step Solution Ch-11  :–

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