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

It is given that the sum of some terms of G.P. is 315 whose first term is 5 and common ratio is 2.

The sum of n terms of the series is,

S n = a( r n −1 ) r−1 315= 5( 2 n −1 ) 2−1 2 n =64 2 n = 2 6

So, the value of n is 6. This means the last term of the G.P. is 6th term.

a n =a r 6−1 =5 ( 2 ) 5 =160

Therefore, 160 is the last term of the G.P.

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