Question

If the A.M. of two positive numbers a and b(a > b) is twice their geometric mean.

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

Answer 1

A.M. between two numbers a and b (a > b) is $\frac{a+b}{2}$

Also, geometric mean between 3 numbers is $\sqrt{ab}$

Given,

A.M = 2 G.M

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

$a+b=4\sqrt{ab}$

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

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

[By componendo and dividendo]

$\frac{(\sqrt{a}+\sqrt{b}{)}^2}{(\sqrt{a}-\sqrt{b}{)}^2=\frac{(\sqrt{3}{)}^2}{(1{)}^2}}$

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

By componendo and dividendo, we get

$\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

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

Squaring both the sides, we get

$\frac{{\left(\sqrt{a}\right)}^2}{{\left(\sqrt{b}\right)}^2}=\frac{{(\sqrt{3}+1)}^2}{{(\sqrt{3}-1)}^2}$

$\Rightarrow \frac{a}{b}=\frac{{(\sqrt{3}+1)}^2}{{(\sqrt{3}-1)}^2}=\frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}$

Taking 2 common from both numerator and denominator

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

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