The sum of some terms of a GP 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 there be $n$ terms in the GP with the first term $a = 5$ and common ratio $r = 2$ .
Then,
Sum of $n$ terms $=$ $315$
$\frac{a (r^{n} - 1)}{r - 1} = 315 \frac{5 (2^{n} - 1)}{2 - 1} = 315 2^{n} - 1 = 63 2^{n} = 64 n = 6$
Therefore, the last term $= a r^{n - 1} = 5 \times 2^{6 - 1} = 160$

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The sum of some terms of a GP is 315, whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Let there be n terms in the GP with the first term a=5 and common ratio r=2. Then,Sum of nterms =315a(rn-1)r-1=31552n-12-1=3152n-1=632n=64n=6Therefore, the last term =arn-1=5×26-1=160

Let there be $n$ terms in the GP with the first term $a = 5$ and common ratio $r = 2$ .
Then,
Sum of $n$ terms $=$ $315$
$\frac{a (r^{n} - 1)}{r - 1} = 315 \frac{5 (2^{n} - 1)}{2 - 1} = 315 2^{n} - 1 = 63 2^{n} = 64 n = 6$
Therefore, the last term $= a r^{n - 1} = 5 \times 2^{6 - 1} = 160$