The sum of two numbers is $6$  times their $G.M$  , then number are in the ratio is

Let two numbers be $a$  and $b$

$G.M=\sqrt{ab}$

$a+b=6\sqrt{ab}\to \left(1\right)$

${\left(a+b\right)}^2=36ab$

${\left(a-b\right)}^2={\left(a+b\right)}^2-4ab$

$=36ab-4ab$

$=32ab$

$a-b=\sqrt{32ab}=4\sqrt{2}\sqrt{ab}\to \left(2\right)$

$\left(1\right)+\left(2\right)$

$2a=\left(6+4\sqrt{2}\right)\sqrt{ab}$

$a=\left(3+2\sqrt{2}\right)\sqrt{ab}$

Sub $a$  in $\left(1\right)$

$b=6\sqrt{ab}-\left(3+2\sqrt{2}\right)\sqrt{ab}$

$=\left(3-2\sqrt{2}\right)\sqrt{ab}$

$\frac{a}{b}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$