Question

anyone help me plzzzz​

Answer 1

Answer:

First term = 108

common ratio = 1/3

Step-by-step explanation:

Given:

The sum of infinite G.P = 162

$\frac{a}{1-r}=162$.........(1)

Also,

The sum of n terms of the G.P = 160

$\frac{a(1-r^n)}{1-r}=160$........(2)

$\frac{(2)}{(1)}\implies$

$\frac{\frac{a(1-r^n)}{1-r}}{\frac{a}{1-r}}=\frac{160}{162}$

$\implies\:1-r^n=\frac{80}{81}$

$\implies\:1-\frac{80}{81}=r^n$

$\implies\:\frac{81-80}{81}=r^n$

$\implies\:\frac{1}{81}=r^n$

$\implies\:r^n=(\frac{1}{3})^4$

$\implies\:r=\frac{1}{3}\:and\:n=4$

$\text{put}\:r=\frac{1}{3}\:\text{in (1)}$

$\frac{a}{1-\frac{1}{3}}=162$

$\implies\:\frac{a}{1-\frac{1}{3}}=162$

$\implies\:\frac{a}{\frac{2}{3}}=162$

$\implies\:\frac{3a}{2}=162$

$\implies\:\frac{a}{2}=54$

$\implies\:\boxed{a=108}$