Question

find the G. P. whose 4th terms is 3/8 and 7th term 3/64.​

Answer 1

$\mathtt{\huge{ \fbox{SOLUTION :}}}$

Given ,

Fourth term of an GP = 3/8

So , $\sf \fbox{a {(r)}^{3} = \frac{3}{8} } - - - (i)$

and

Seventh term of an GP = 3/64

So , $\sf \fbox{a {(r)}^{6} = \frac{3}{8} } - - - (ii)$

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

$\sf \hookrightarrow {(r)}^{3} = \frac{3 \times 8}{64 \times 3} \\ \\ \sf \hookrightarrow {(r)}^{3} = \frac{1}{8}$

Substitute the value of (r)³ in equation (i) , we get

$\sf \hookrightarrow a \times \frac{1}{8} = \frac{3}{8} \\ \\ \sf \hookrightarrow a = 3$

Hence , the common ratio and first term of GP are 1/2 and 3

We know that , The GP can be written as

$\large \fbox{a \:, \: ar \:, \: a {r}^{2} \: , \: ..... \: \: a {r}^{n - 1} }$

Therefore , the required GP will be 3 , 3/2 , 3/4 , 3/8 , .......