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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, S_{n }`(a(r^n - 1))/(r - 1)`

given geometric progression

5 + 10 + 20 + 40 + …….

Sum of n terms = `(5(2^"n" - 1))/(2 -1) = 315`

∴ 2^{n} – 1 = 63

or 2^{n} = 64 = 2^{6}

n = 6

6th term = 5 × 2^{6} – 1

= 5 × 2^{5}

= 5 × 32

= 160

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