Question

If the A.M. and G.M between two numbers are in the ration m : n then the numbers are in the ratio

• A. $m+\sqrt{n^2-m^2}:m-\sqrt{n^2-m^2}$
• B. $m+\sqrt{m^2+n^2}:m\sqrt{m^2+n^2}$
• C. $m+\sqrt{m^2-n^2}:m-\sqrt{m^2-n^2}$
• D. $n+\sqrt{m^2-n^2}:n-\sqrt{m^2-n^2}$

Answer 1

The correct option is C $m+\sqrt{m^2-n^2}:m-\sqrt{m^2-n^2}$

Let Two Numbers be $a\&b$

Then $AM=\frac{a+b}{2},Gm=\sqrt{ab}$

Given $\frac{AM}{Gm}=\frac{m}{n}$

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

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

Applying componendo and dividendo (CD rule)

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

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

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

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

Again applying CD rule

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

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

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