Question

Find the sum of the geometric series 1+$\sqrt{3}$+3...+81.

Answer 1

Given

• The geometric series $1, \sqrt{3} , 3 ..........81$

To find

Sum of the geometric series

Formula

Let a be the first term and r be the common ratio of a G.P., then sum of n terms is given by the formula

$\boxed{ \: \green{ \frac{a( {r}^{n } - 1)}{r - 1} } \: }$

Solution

Given , G.P. = 1, $\sqrt{3}$ , 3, ......81

Here, a = 1 and r = $\frac{ \sqrt{3} }{1} }$

Also, since 81 is the last lerm . Using l = a ${r}^{n - 1}$

$81 = 1 {( \sqrt{3} )}^{n - 1} \\ \\ = > ( { \sqrt{3} )}^{8} = {( \sqrt{3} )}^{n - 1} \\ \\ = > n - 1 = 8 \\ \\ = > n = 9$

There are total 9 terms.

Hence, sum of G.P. =

$\frac{a( {r}^{n } - 1)}{r - 1} \\ \\ = > \frac{1( { \sqrt{3}) }^{9} - 1}{ \sqrt{3} - 1 } \\ \\ = > \frac{81 \sqrt{3} - 1 }{ \sqrt{3} - 1 }$