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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 86 of 106
Marks: +1, -0
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Solution:  
We have, a = 5, r = 2
Let the number of terms be n whose sum is 315.
Then a(4n−1)r−1\frac{a(4^n-1)}{r-1} = 315 ⇒ 5(2n−1)2−1\frac{5(2^n-1)}{2-1} = 315
⇒ 2n−12^n - 1 = 63 ⇒ 2n2^n = 64 ⇒ 2n2^n = (2)6(2)^6
∴ n = 6
Hence, the last term = arn−1ar^{n-1} i.e., a6a_6 = 5(2)6−15(2)^{6-1} = 5(2)55(2)^5 = 160
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