# The ratio of the A.M and G.M. of two positive numbers a and b,

Question:

The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that $a: b=\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)$

Solution:

Let the two numbers be a and b.

A.M $=\frac{a+b}{2}$ and G.M. $=\sqrt{a b}$

According to the given condition,

$\frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}$

$\Rightarrow \frac{(a+b)^{2}}{4(a b)}=\frac{m^{2}}{n^{2}}$

$\Rightarrow(a+b)^{2}=\frac{4 a b m^{2}}{n^{2}}$

$\Rightarrow(a+b)=\frac{2 \sqrt{a b} m}{n}$ $\ldots(1)$

Using this in the identity $(a-b)^{2}=(a+b)^{2}-4 a b$, we obtain

$(a-b)^{2}=\frac{4 a b m^{2}}{n^{2}}-4 a b=\frac{4 a b\left(m^{2}-n^{2}\right)}{n^{2}}$

$\Rightarrow(a-b)=\frac{2 \sqrt{a b} \sqrt{m^{2}-n^{2}}}{n}$ $\ldots(2)$

Adding $(1)$ and $(2)$, we obtain

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

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

Substituting the value of a in (1), we obtain

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

$=\frac{\sqrt{a b}}{n} m-\frac{\sqrt{a b}}{n} \sqrt{m^{2}-n^{2}}$

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

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

Thus, $a: b=\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)$