The product of three numbers in G.P. is 216. If 2, 8, 6 be added to them,

Solution:

Let the terms of the given G.P. be $\frac{a}{r}, a$ and $a r$.

∴ Product = 216

$\Rightarrow a^{3}=216$

$\Rightarrow a=6$

It is given that $\frac{a}{r}+2, a+8$ and $a r+6$ are in A.P.

$\therefore 2(a+8)=\frac{a}{r}+2+a r+6$

Putting $a=6$, we get

$\Rightarrow 28=\frac{6}{r}+2+6 r+6$

$\Rightarrow 28 r=6 r^{2}+8 r+6$

$\Rightarrow 6 r^{2}-20 r+6=0$

$\Rightarrow(6 r-2)(r-3)=0$

$\Rightarrow r=\frac{1}{3}, 3$

Hence, putting the values of $a$ and $r$, the required numbers are $18,6,2$ or 2,6 and 18 .