Question

The third term is 2 and 7th is 162 of GP.Find the sum of first ten terms

Answer 1

Given ,

The third term of GP is 2

The seventh term of GP us 162

We know that , the nth term of GP is given by

$\boxed{ \tt{a_{n} = a {(r)}^{n - 1} }}$

Thus ,

$\sf a {(r)}^{2} = 2 - - - (i)$

and

$\sf a {(r)}^{6} = 162 - - - (ii)$

Divide eq (ii) by (i) , we get

$: \mapsto \tt \frac{a {(r)}^{2} }{a {(r)}^{6} } = \frac{162}{2}$

$: \mapsto \tt {(r)}^{6 - 2} = 81$

$: \mapsto \tt {(r)}^{4} = 81$

$: \mapsto \tt r = \sqrt[4]{81}$

$: \mapsto \tt r = 3$

Put r = 3 in eq (i) , we get

$: \mapsto \tt a {(3)}^{2} = 2$

$: \mapsto \tt a = \frac{2}{9}$

Now ,

The sum of first n term of GP is given by

$\boxed{ \sf{ s_{n}= \frac{a( {r}^{n} - 1)}{r - 1} }}$

Thus ,

$\hookrightarrow \tt s_{10} = \frac{2}{9} \times \frac{ ({(3)}^{10} - 1)}{(3 - 1)}$

$\hookrightarrow \tt s_{10} = \frac{59049 - 1}{9}$

$\hookrightarrow \tt s_{10} =6560.88$