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Question
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.
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Solution
Let the sum of n terms of the G.P. be 315.
It is known that, Sn `(a(r^n - 1))/(r - 1)`
given geometric progression
5 + 10 + 20 + 40 + …….
Sum of n terms = `(5(2^"n" - 1))/(2 -1) = 315`
∴ 2n – 1 = 63
or 2n = 64 = 26
n = 6
6th term = 5 × 26 – 1
= 5 × 25
= 5 × 32
= 160
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