Question

If the ratio of A.M between two positive real numbers $a$ and $b$ to their H.M is $m:n$, then $a:b$ is

• A. $\frac{(\sqrt{m}-\sqrt{n}+\sqrt{n})}{(\sqrt{(m-n)}-\sqrt{n})}$
• B. $\frac{(\sqrt{n}+\sqrt{(m-n)})}{(\sqrt{n}-\sqrt{(m-n)})}$
• C. $\frac{(\sqrt{m}+\sqrt{(m-n)})}{(\sqrt{m}-\sqrt{(m-n)})}$
• D. $noneofthese$

Answer 1

The correct option is C

$\frac{(\sqrt{m}+\sqrt{(m-n)})}{(\sqrt{m}-\sqrt{(m-n)})}$

Explanation for correct option :

Step 1. Use the concept of arithmetic and harmonic mean between any two numbers

We know that AM between any two numbers $a$ and $b$ is given by $AM=\frac{(a+b)}{2}$

Also, the GM between any two numbers $a$ and $b$ is given by $HM=\frac{2ab}{(a+b)}$

It is given that the ratio of A.M to the GM between any two real numbers $a$ and $b$ is $m:n$

$\therefore \frac{AM}{HM}=\frac{m}{n}$

$\Rightarrow$$\frac{(a+b)}{2}÷\frac{2ab}{(a+b)}=\frac{m}{n}$

$\Rightarrow$$\frac{(a+b)}{2}\times \frac{(a+b)}{2ab}=\frac{m}{n}$

$\Rightarrow$ $\frac{{(a+b)}^2}{4ab}=\frac{m}{n}\dots \left(i\right)$

$\Rightarrow$ $\left[\frac{{(a+b)}^2}{4ab}\right]–1=\frac{m}{n}–1$ [ Subtract $1$ from both sides]

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

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

Dividing equation first by equation second, we get

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

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

Step 2: Applying componendo and dividendo rule

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

$\Rightarrow$ $\frac{2a}{2b}=\frac{[\sqrt{m}+\sqrt{(m-n)}]}{[\sqrt{m}-\sqrt{(m-n)}]}$

$\Rightarrow$ $\frac{a}{b}=\frac{[\sqrt{m}+\sqrt{(m-n)}]}{[\sqrt{m}-\sqrt{(m-n)}]}$

Thus, Option (C) is correct.