Question

find the sum of the given series by geomatric prograssion..​

Answer 1

$\mathtt{ \huge\fbox{SOLUTION :</p><p>}}$

Given ,

First term (a) = 1/8

Common ratio (r) = 2

nth term (an) = 64

We know that , the nth term of an GP is given By

$\sf \large \fbox{a_{n} = a {r}^{(n - 1)} }$

Substitute the known values , we obtain

$\sf \mapsto 64 = \frac{1}{8} {(2)}^{(n - 1)} \\ \\ \sf \mapsto 512 = {(2)}^{(n - 1)} \\ \\ \sf \mapsto {(2)}^{9} ={(2)}^{(n - 1)}$

By comparing the powers , we get

$\sf \mapsto 9 = n - 1 \\ \\ </p><p> \sf \mapsto n = 10$

Hence , 10th term of GP is 64

We know that , when common ratio > 1 , the sum of first n terms of GP is given by

$\sf \star \: \: \: \: \: \: \: \: \: \: \fbox{ S = \frac{a( {r}^{n} - 1)}{ r- 1}} \\ \\ \sf \mapsto S = \frac{ \frac{1}{8} ({(2)}^{10} - 1) }{2 - 1} \\ \\\sf \mapsto S = \frac{ \frac{1}{8}(1024 - 1) }{1} \\ \\ \sf \mapsto S =\frac{1}{8} \times 1023 \\ \\\sf \mapsto S = 127.875$

Hence , 127.875 is the sum of given GP