Question

If the AM of $2$ numbers $a$ and $b$ is equal to $\sqrt{10}$ times their $GM$, then $\frac{(a-b)}{(a+b)}$ =

• A. $\frac{\sqrt{10}}{3}$
• B. $3\sqrt{10}$
• C. $\frac{3}{\sqrt{10}}$
• D. $\frac{9}{10}$

Answer 1

The correct option is C

$\frac{3}{\sqrt{10}}$

Explanation for Correct Option :

Step 1: Apply the formula

Let three numbers is $a,b,c$

Arithmetic Mean $=\frac{a+b}{2}$

Geometric Mean $=\sqrt{ab}$

Apply the above formula in the below equation

$\mathrm{A}.\mathrm{M}.=\sqrt{10}(\mathrm{G}.\mathrm{M}.)$

$$\begin{gathered} \Rightarrow \frac{(a+b)}{2}=\sqrt{10}\sqrt{\left(ab\right)} \\ \Rightarrow \frac{{(a+b)}^2}{4}=10ab \\ {\Rightarrow (a+b)}^2=40ab \\ \Rightarrow a^2+2ab+b^2=40ab \\ \Rightarrow a^2+b^2=38ab \\ \Rightarrow \frac{\left(a^2+b^2\right)}{2ab}=19.....\left(\mathrm{i}\right) \end{gathered}$$

Step 2: Add $-2\mathrm{ab}$ in equation (i)

$\Rightarrow a^2+b^2-2ab=38ab-2ab$

Now,

$$\begin{gathered} \Rightarrow \frac{(a^2+b^2-2ab)}{(a^2+2ab+b^2)}=\frac{36}{40} \\ \Rightarrow \frac{(a^2+b^2-2ab)}{(a^2+2ab+b^2)}=\frac{9}{10} \\ \Rightarrow \frac{{(a-b)}^2}{{(a+b)}^2}=\frac{9}{10} \\ \Rightarrow \frac{(a-b)}{(a+b)}=\frac{3}{\sqrt{10}} \end{gathered}$$

Hence option (4) is the answer.