Mathematics - Sequence & Series Question with Solution | TestHub

To minimize the calculation, $3$ numbers in $G.P.$ can be taken as $\frac{a}{r},a,ar$ where, $a$ is the first term and $r$ is the common ratio.

Given product of three numbers is $512$.

$\Rightarrow \frac{a}{r}.a.ar=512\Rightarrow a^3=8^3\Rightarrow a=8$

Also given, $\frac{a}{r}+4,a+4,ar$ are in

$A.P.$

$\Rightarrow 2\left(a+4\right)=\frac{a}{r}+4+ar$

Putting $a=8$, we get

$2\left(12\right)=\frac{8}{r}+4+8r$

$\Rightarrow 2r^2-5r+2=0$

$\Rightarrow \left(2r-1\right)\left(r-2\right)=0$

$\Rightarrow r=2$ or $\frac{1}{2}$

Hence, numbers may be $4,8,16$ or $16,8,4$

In both the cases, sum $=4+8+16=28$.