Question

The product of three terms of a $GP$ is $512$. If $8$ is added to the first term and $6$ to the second term, find the new terms of an $GP$.

Answer 1

$\to$ put $3$ terms of $G.P$ be $\frac{a}{r},a,ar$

Given :- product of $3$ terms of $G.P.=512$

So, $\left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=a^3=512$

$a^3=(2{)}^9$

$a=8$

when $8$ is added to $I^{st}$ term and $6$ to second term.

it series is still in $G.P.$

New series $(\frac{a}{r}+8),(a+b),ar$

as in $G.P.$:- $(\frac{a}{r}+8)\left(ar\right)=(a+b)$

$a^2+8ar=a+b$

from $\left(1\right)$ :-

$64+64r=8+b$

$64r=14-64\Rightarrow r=\frac{-50}{64}$

$r=\frac{-25}{32}$

New $G.P.$ terms :- $\frac{8}{\frac{-25}{32}}+8,8+6,8.\left(\frac{-25}{32}\right)$

$=\frac{56}{25},14,-\frac{25}{4}$