# ICSE Solutions for Chapter 11 Geometric Progression Class 10 Mathematics

Question 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, …

Solution 1:
(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) = 1/3, (1/72)/(1/24) = 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.

Question 2:
Find the 9th term of the series :
1, 4, 16, 64 ...

Solution 2:
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 first term, a = 1 and common ratio, r = 4
Now, tn = arn - 1
⇒ t9 = 1×48 = 65536

Question 3:
Find the seventh term of the G.P. :
1, √3, 3, 3√3

Solution 3:
Given G.P.:  1, √3, 3, 3√3,…
Here,
First term, a = 1
Common ratio, r = √3/1 = √3
Now, tn = arn - 1
⇒ t7 = 1×(√3)6 = 27

Question 4:
Find the 8th term of the sequence :
3/4, 1 1/2, 3, …….

Solution 4:
Given sequence: 3/4., 1½, 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 = ¾  and common ratio, r = 2
Now, tn = arn - 1
⇒  t8 = ¾ × 27 = ¾ × 2 × 2 × 2 × 2 × 2 × 2 × 2  = 3 × 25 = 96

Question 5:
Find the 10th term of the G.P. :

Solution 5:
Given G.P.: 12, 4, 1 1/3,…….
Here,
First term, a = 12
Common ratio, r = 4/12 = 1/3
Now, tn = arn - 1
⇒ t10 = 12 × (1/3)9 = 12 × 1/19683  = 4/6561

Question 6:
Find the nth term of the series :

Solution 6:
Given series: 1, 2, 4, 8, ……
Now,
2/1 = 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, tn = arn - 1
⇒ tn = 1 × 2n - 1 = 2n - 1

Question 7:
Find the next three terms of the sequence :
√5, 5, 5√5, …

Solution 7:
Given sequence: √5, 5, 5√5,……
Now,
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, tn = arn - 1
∴ Next three terms :
4th term = √5×(√5)3 = √5×5√5 = 25
5th term = √5×(√5)4 = √5×25 = 25√5
6th term = √5×(√5)5 = √5×25√5 = 125

Question 8:
Find the sixth term of the series :
22, 23, 24, …

Solution 8:
Given  sequence : 22, 23, 24,…..
Now, 23/22 = 2,  24/23 = 2
Since 23/22 = 24/23 = ……= 2, the given sequence is a G.P.
With first term, a = 22 = 4 and common ratio, r = 2.
Now, tn = arn - 1
∴ t6 = 4 × (2)5 = 4×32 = 128

Question 9:
Find the seventh term of the G.P. :
3 + 1, 1, (√3 – 1)/2, …

Solution 9:
Given G.P.: √3 + 1, 1, (√3 – 1)/2, …
Here,
First term, a = √3 + 1
Common ratio, r = 1/(√3 + 1)
Now, tn = arn - 1
⇒ t7 = {(√3 + 1)×(1/(√3 + 1)}6
= (1/(√3 + 1)5
= {1/(√3 + 1) × (√3 – 1)/(√3 – 1)}5
= {(√3 – 1)/2}5
= 1/32(√3 – 1)5

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

Solution 10:
First term , a = 64
Second term, t2 = 32
⇒ ar = 32
⇒ 64 × r = 32
⇒ r = 32/64 = ½
∴  Required  G.P. = a, ar, arn-1, arn-2, …
= 64, 32, 64 × (1/2)2, 64 × (1/2)3,…
= 64, 32, 16, 8, ……