Question

The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$.

Answer 1

Let a, b be the numbers

Let the geometric mean between them be G we have,

$a+b=6G$ (G.M of a, b)

But G = $\sqrt{ab}$

$\therefore a+b=6\sqrt{ab}$

$\Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac{3}{1}$

Applying components and dividents,

$\Rightarrow \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{3+1}{3-1}$

$\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{2}}{1}$

Again applying components and dividends,

$\Rightarrow \frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$

$\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$

$\Rightarrow \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$

Squaring both the sides,

${\left(\frac{\sqrt{a}}{\sqrt{b}}\right)}^2=\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}-1{)}^2}$

$\Rightarrow \left(\frac{a}{b}\right)={\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)}^2$

$\Rightarrow \left(\frac{a}{b}\right)=\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}}$

$\Rightarrow \left(\frac{a}{b}\right)=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$

$a:b=(3+2\sqrt{2}):(3-2\sqrt{2})$