Question

if the A.M. between a and b is equal to n times their G.M
find the ratio of a to b

Answer 1

HELLO DEAR,

Given A.M. between two positive number a and b is n times the G.M.

⇒ A.M. = n G.M.

(a + b)/2 = n√ab

(a + b)/2√ab = n/1

Applying componendo and Dividendo

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

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

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

again, applying componendo and dividend.

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

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

$\bold{\frac{a}{b} = \frac{n + 1 + n - 1 + 2\sqrt{(n + 1)(n - 1)}}{n + 1 - n + 1 - 2\sqrt{(n + 1)(n - 1)}}}$

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

$\bold{\frac{a}{b} = \frac{n + \sqrt{n^2 - 1}}{1 - \sqrt{n^2 - 1}}}$

Hence, the ratio of a to b is $\bold{[n + \sqrt{n^2 - 1}]}$ : $\bold{[1 - \sqrt{n^2 - 1}]}$

I HOPE ITS HELP YOU DEAR,
THANKS