Question

The product of three numbers in $G.P.$ is $216.$ If $2,8,6$ be added to them$,$ the results are in $A.P.$ Find the numbers$.$

Answer 1

Let the numbers be $:$

$\frac{a}{r},a,ar$

It is given that product of three numbers in $G.P.$ is $216.$

$\frac{a}{r}\times a\times ar=216$

$\Rightarrow a^3=216$

$\Rightarrow a^3=6^3$

$\Rightarrow a=6$

Now $2,8,6$ be added to them then the number becomes

$\frac{a}{r}+2,a+8,ar+6$ are in $A.P.$

$\frac{6}{r}+2,6+8,6r+6$ are in $A.P.$

$\Rightarrow 14-\frac{6}{r}-2=6r+6-14$

$\Rightarrow 12-\frac{6}{r}=6r-8$

$\Rightarrow 3r^2-10r+3=0$

$\Rightarrow \left(r-3\right)\left(r-\frac{1}{3}\right)=0$

$r=3orr=\frac{1}{3}$

when $r=3$ then numbers are

$2,6,18$

When $r=\frac{1}{3}$ then numbers are

$18,6,2$

Hence, the numbers are $2,6,18$ $or$ $18,6,2$