Selina Concise Class 10 Math Chapter 11 Exercise 11A Geometric Progression Solutions
Geometric Progression
Exercise: 11A
Question: 1
Find which of the following sequence form a Geometric progression
(1) 8, 24, 72, 216, ——–
= Solution: Given sequence id 8, 24, 72, 216 ——–
Here, a1 = 8, a2 = 24, a2 = 72
a2/a1 = 24/8 = 3, a3/a2 = 72/24 = 3, ————
Therefore, a2/a1 = a3/a2 = 3
Therefore, The given sequence is in geometric progression with common ratio is 3.
“If the given sequence have common ratio then this sequence is in geometric progression.
(ii) 1/8, 1/24, 1/72, 1/216,
= Solution: Given sequence is 1/8, 1/24, 1/72, 1/216, ——–
Here, a1 = 1/8, a2 = 1/24, a3 = 1/72 ————–
Therefore, a2/a1 = (1/24)/(1/8), =1/24 x 8/1 = 1/3
a3/a2 =( 1/72)/1/24 = 1/72 X 24/1 = 1/3 ————
Therefore, a2/a1 = a3/a2 = ——- = 1/3
Therefore, The given sequence is in geometric progression, with common ratio is 1/3
“If the given sequence Have common ratio then this sequence is in geometric progression”
(iii) 9, 12, 16, 24, —–
= Solution: Given sequence is 9, 12, 16, 24, —–
Here, a1 = 9, a2 = 12, a3 = 16, ——
a2/a1 = 12/9 = 4/3
A3/a2 = 16/12 = 8/6 = 4/3
Therefore, a2/a1 = a3/a2 = ——-= 4/3
Therefore, The given sequence is in geometric progression, with common ratio 4/3.
“If the given sequence have common ratio then this sequence is in Geometric progression.”
Question: (2)
Find the 9th term of the series 1, 4, 16, 64, ——-
= Solution:
1, 4, 16, 64, ———
Here, a1 = 1, a2 = 4, a3 = 16, ——
t1 = a = 1
We have to find common ratio
a2/a1 = 4/1 = 4
a3/a2 = 16/4 = 4
Therefore, a2/a1 = a3/a2 = —— = 4
Therefore, The common ratio (r) = 4
Now, We have to find 9th term
We know that,
tn = arn-1
Therefore, t9 = (1) (4)9-1
= 48 (Because, n = 9)
T9 = 65536
Therefore, The 9th term is 65536
Question (3):
Find the seventh term of the geometric progression:
1, √3, 3, 3 √3, —–
= Solution:- The given sequence is 1, √3, 3, 3 √3, ——
Here, t1 = a = 1
Now, We have to find common ratio
a1 = 1, a2 = √3, a3 = 3, ——
a2/a1 = √3/1
a3/a2 = 3/ √3 = √3 x √3/ √3 = √3
a2/a1 = a3/a2 = ——– = √3
Therefore, The common ratio (r) = √3
We know that,
tn = arn-1
we have to find 7th term
t7 = 1(√3) 7-1 (n = 7)
= (√3)6
= √3 x √3 x √3 x √3 x √3 x √3
T7 = 27
Question 4:
Find the 8th term of the sequence:
3/4, 1 1/2, 3, —–
= Solution:-
Given sequence is 3/4, 1 1/2, 3, ——
Here, a1 = 3/4, a2 = 11/2 = 2+1/2 = 3/2, a3 = 3, —-
Therefore, t1 = a = 3/4
Now, We have to find common ratio
a2/a1 = 3/2/3/4 = 3/2 x 4/3 = 2/1
a3/a2 = 3/3/2 = 3 x 2/3 = 2
a2/a1 = a3/a2 = ——– = 2
Therefore, Common ratio (r) = 2
Now, we have to find 8th term
We know that,
tn = arn-1
t8 = arn-1
t8 = (3/4) (2)8-1 (n = 8)
= 3/4 (2)7
= 3/4 (2 x 2 x 2 x 2 x 2 x 2 x 2)
= 3 x (2 x 2 x 2 x 2 x 2)
= 3 x (32)
T8 = 96
Therefore, The 8th term is 96.
Question: (5)
Find the 10th term of the geometric progression:
12, 4, 1 1/3, —–
= Solution:
Given sequence is 12, 4, 1 1/3, —–
Here, a1 = 12, a2 = 4, a3 = 1 1/3 = 3+1/3
= 4/3
a3 = 4/3
Therefore, t1 = a1 = 12
Now, We have to find common ratio (r)
a2/a1 = 4/12 = 1/3
a3/a2 = 413/4 = 4/3 x 1/4 = 1/3
Therefore, a2/a1 = a3/a2 = ——- = 1/3
Therefore, Common ratio (r) = 1/3
Now, we have to find 10th term
we know that,
tn = ar n-1
t7 = (12) (1/3)10-1
= 12 (1/3)9
= 12 x 1/3 x 1/3 x 1/3 x 1/3 x 1/3 x 1/3 x 1/3 x 1/3 x 1/3
= 4 x 1/6561
T7 = 4/6561
Therefore, The 7th term is 4/6561
Question (6):
Find the nth term of the series: 1, 2, 4, 8, ——-
= Solution:-
The given sequence is 1, 2, 4, 8, ——
Here, a1 = 1, a2 = 21, a3 = 4, ——
Now, we have to find common ratio
a2/a1 = 2/1 = 2
a3/a2 = 4/2 = 2
a2/a1 = a3/a2 = ——– = 2
Therefore, r = 2
Now, We have to find nth term, we know that,
tn = ar n-1
tn = 1(2) n-1
tn = 2n – 2
Therefore, The nth term is 2n – 2
Here is your solution of Selina Concise Class 10 Math Chapter 11 Exercise 11A Geometric Progression.
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