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

Geometric Progression Exe-11C Class-10 Concise ICSE Maths Solution Ch-11.  In this article you would learn to solve proving questions 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-11C 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-11C	proving questions on general term of G. P.
Edition	2025-2026

proving questions on general term of G. P.

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

Que-1: Find the seventh term from the end of the series :
√2, 2, 2√2,……32

Ans: √2,2,2√2,………,32.
Now, 2√2=2√2/2=√2
So, the given series is a G.P. with common ratio, r = √2
Here, last term, l = 32
∴ 7^th term from an end = 1/r^6
= 32/(√2)^6
= 32/8
= 4

Que-2: Find the third term from the end of the GP.
2/27, 2/9, 2/3, ………., 162

Ans: Given G.P. : 2/27,2/9,2/3,………,162
Common ratio, r =2/9 / 2/27 = 3
Last term, l = 162
∴ 3^rd term from an end =1r2
= 162/(3)^2
= 162/9
= 18

Que-3: For the G.P. 1/27, 1/9, 1/3, ……, 81; find the product of fourth term from the beginning and the fourth term from the end.

Ans: Here,
Common ratio, r = 1/9 / 1/27 = 3
First term, a = 1/27 and last term, l = 81
∴ 4^th term from the beginning = ar³
= 1/27×(3)³
= 1/27×27
= 1
And 4^th term from an end = 1/r³
= 81/(3)³
= 81/27
= 3
Thus, required product = 1 × 3 = 3

Que-4: If for a G.P., pth, qth and rth terms are a, b and c respectively ;
prove that : (q – r) log a + (r – p) log b + (p – q) log c = 0

Ans: Let the first term of the G.P. be a and its common ratio be R.

Then,
p^th term = a => AR^{p – 1} = a
q^th term = b => AR^{q – 1} = b
r^th term = c => AR^r – 1 = c
Now,
a^{q – r} × b^{r – p} × c^{p – q} = (AR^{p – 1})^{q – r} × (AR^{q – 1})^{r – p} × (AR^{r – 1})^{p – q}
= Aq-r.R(p-1)(q-r)×Ar-p.R(q-1)(r-p)×Ap-q.R(r-1)(p-q)
= Aq-r+r-p+p-q×R(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)
= A⁰ × R⁰
= 1
Taking log on both the sides, we get
log (a^{q – r} × b^{r – p} × c^{p – q}) = log 1
=> (q – r) log a + (r – p) log b + (p – q) log c = 0    …(proved)

Que-5: If a, b and c in G.P., prove that : log an, log bn and log cn are in A.P.

Ans: Here, a, b, c are in G.P.
b² = ac
Taking log on both sides, we get
log(b²) = log(ac)
2log b = log a + log c
log b + log b = log a + log c
log b – log a = log c – log b
log a, log b and log c are in A.P.

Que-6: If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.

Ans: Let a, b, c are in G.P.
Then b2 = ac …(i)
Now ax, bx + cx will be in G.P. if (bx)2 = ax.cx
=> (bx)2 = ax.cx
=>(b2)x = (ac)x
Hence ax, bx, cx are in G.P. (∴ b2 = ac)
Hence proved.

Que-7: If a, b and c are in A.P. a, x, b are in G.P. whereas b, y and c are also in G.P. Show that : x², b², y² are in A.P.

Ans: a, b and c are in A.P.
=> 2b = a + c
a, x and b are in G.P.
=> x² = ab
b, y and c are in G.P.
=> y² = bc
Now,
x² + y² = ab + bc
= b(a + c)
= b × 2b
= 2b²
=> x², b² and y² are in A.P.

Que-8: If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that :
(i)1/x + 1/y = 2/b
(ii)a/x + c/y = 2

Ans:
(i)
a, b and c are in G.P.
=> b² = ac
a, x, b, y and c are in A.P.
=> 2x = a + b => x=a+b/2
2b = x + y => b=x+y/2
2y = b + c => y=b+c/2
Now,
1/x+1/y=2/a+b+2/b+c
= 2b+2c+2a+2b/ab+ac+b2+bc
= 2a+2c+4b/ab+b2+b2+bc
= 2a+2c+4b/ab+2b2+bc
= 2(a+c+2b)/b(a+2b+c)
= 2/b
(ii)
a, b and c are in G.P.
=> b² = ac
a, x, b, y and c are in A.P.
=> 2x = a + b => x=a+b/2
2b = x + y => b=x+y/2
2y = b + c => y=b+c/2
Now,
a/x+c/y=2a/a+b+2c/b+c
= 2a(b+c)+2c(a+b)/(a+b)(b+c)
= 2ab+2ac+2ac+2bc/ab+ac+b2+bc
= 2ab+4ac+2bc/ab+b2+b2+bc
= 2(ab+2ac+bc)/ab+2b2+bc
= 2(ab+2ac+bc)/ab+2ac+bc
= 2

Que-9: If a, b and c are in A.P. and also in G.P., show that: a = b = c.

Ans: a, b and c are in A.P.
2b = a + c
b=a+c/2
a, b and c are also in G.P.
b² = ac
(a+c)²/2=ac
a²+c²+2ac= 4ac
a² + c² + 2ac = 4ac
a² + c² – 2ac = 0
(a – c)² = 0
a – c = 0
a = c
Now, 2b = a + c
2b = a + a
2b = 2a
b = a
Thus, we have a = b = c

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