Concept:

If A is the arithmetic mean of numbers a and b and is given by ⇔ A =  (a + b)/2

If G is the geometric mean of the numbers a and b and is given by  ⇔ G = \(\rm \sqrt{ab}\)

Calculation:

Let the numbers be a and b 

According to given condition,

\(\rm \frac{a+b}{2\sqrt {ab}}=\frac m n\)

Applying componendo and dividendo, we get 

\(\begin{array}{l} \rm \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{m+n}{m-n} \\ \rm \frac{(\sqrt{a}+\sqrt{b})^{2}}{(\sqrt{a}-\sqrt{b})^{2}}=\frac{m+n}{m-n} \\ \rm \frac{(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}}{\sqrt{m-n}} \end{array}\)

Applying componendo and dividendo again, we get 

\( \rm \frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \)

Squaring both sides, we get

\(\rm \frac{a}{b}=\left(\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}\right)^{2} \)

\(\rm\frac{a}{b}=\frac{2 m+\sqrt{m^{2}-n^{2}}}{2 m-2 \sqrt{m^{2}-n^{2}}} \\ \rm \frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)