Selina Concise Class 10 Math Chapter 11 Exercise 11C Geometric Progression Solutions
Geometric Progression
Exercise: 11C
Question: 1
Find the seventh term from the end of the series: √2, 2, 2, √2, —– 32
= Solution:
√2 , 2, 2√2, ——–, 32
Here, a = √2
Common ratio (r) = r2/r1 = 2/√2
= √2 x √2/√2
R = √2
And the last term tn = 32
We know that,
tn = arn-1
32 = √2 x (√2)n-1
32 = (√2)1 x (√2)n-1
= (√2)1 + n-1
32 = (√2)n
But the power of 2 is 5
(2)5 = 32
Therefore,
(2)5 = (√2)n
(√2)5 x 2 = (√2)n
(√2)10 = (√2)n
(√2)10 = (√2)n
The base is same
Now, equating the powers or exports
10 = n
Therefore, n = 10
We have find 7th term from the end is
(n – n + 1)th tern
Therefore, (10 – 7 + 1)th
Therefore, (10 – 7 + 1)th
= (3 + 1)th
= 4th term
i.e. we have to find 4th term of the geometric progression
Therefore, t4 = ar4-1
= ar3
= (√2) (√2)3
= (√2) x (√2 x √2 x √2)
t4 = (√2) x (2√2)
t4 = √2 x 2√2
= √2 x √2 x2
t4 = 2 x 2
t4 = 4
Question 2:
Find the third term from the end of the geometric progression
2/27, 2/9, 2/3, ——, 162
=> Solution:- Given series
2/27, 2/9, 2/3, ——, 162
Here, t1 = a = 2/27
Common ratio (r) = r2/r1
= 2/9/2/27
= 2/9 x 27/2
r = 3
Also, tn = 162
We know that,
tn = arn-1
162 = arn-1
162 = 2/27 (3)n-1
162 x 27/2 = (3)n-1
81 x 27 = (3)n-1
2187 = (3)n-1
Now, take, The power of 3 is 7 so (37) = 2187
(3)7 = (3)n-1
The base is same, so, equating the powers or exponents,
7 = n-1
7+1 = n
Therefore, n = 8
We have to find third term from the end, (8-3+1)th = (5+1) = 6th term
Therefore,
t6 = arn-1
= ar6-1
= 2/27 (3)6-1
= 2/27 (3)5
t6 = 2/(3)3 (3)5
t6 = 2 x (3)5 x (3)-3
= 2 x (3)5-3
= 2(3)2
t6 = 2 x 9
t6 = 18
Question: 3
Find the geometric progression 1/27, 1/9, 1/3, —–, 81 find the product of fourth term from the beginning and the fourth terms from the end
Solution:-
Given geometric progression
1/27, 1/9, 1/3, —–, 81
Here, t1 = a = 1
Common ratio (r) = r2/r1
r = 1/9/1/27 = 1/9 x 27/1 = 3
r = 3
And the last term (tn) = 81
We know that,
tn = ar n-1
81 = 1/27 x (3)n-1
81 x 27 = (3)n-1
2187 = (3)n-1
(3)7 = (3)n-1
The base is same, so equating the exponents,
7 = n-1
7 + 1 = n
n = 8
Therefore, Total 8 terms in the given geometric progression.
Now, we have to find 4th term from the beginning and the fourth them from the end — (8 – 4 + 1) = 5th term
Now, we have to find product of t4 and t5,
t4 = ar4-1 = ar3 and
t5 = ar5-1 = ar4
Here is your solution of Selina Concise Class 10 Math Chapter 11 Exercise 11C Geometric Progression
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