If both the A.M. between m and n and G.M. between two distinct positive numbers a and b are equal to  

$\frac{ma+nb}{m+n}$, then n is equal to 

$\frac{ma+nb}{m+n}=\frac{1}{2}(m+n)=\sqrt{ab}$                     (1)

$\Rightarrow ma+nb=\sqrt{ab}(m+n)$

$\Rightarrow m\sqrt{a}(\sqrt{a}-\sqrt{b})=n\sqrt{b}(\sqrt{a}-\sqrt{b})$

$\Rightarrow \frac{m}{\sqrt{b}}=\frac{n}{\sqrt{a}}=k\left(\text{ say }\right)$                          (2)

From (1) and (2) 

$\frac{1}{2}k(\sqrt{a}+\sqrt{b})=\sqrt{ab}\Rightarrow k=\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$

Thus, $n=k\sqrt{a}=\frac{2a\sqrt{b}}{\sqrt{a}+\sqrt{b}}$