Question

The $A.M$. between $m$ and $n$ and the $G.M$. between $a$ and $b$ are each equal to $\frac{ma+nb}{m+n}$. Then $m=$

• A. $\frac{a\sqrt{b}}{\sqrt{a}+\sqrt{b}}$
• B. $\frac{b\sqrt{a}}{\sqrt{a}+\sqrt{b}}$
• C. $\frac{2a\sqrt{b}}{\sqrt{a}+\sqrt{b}}$
• D. $\frac{2b\sqrt{a}}{\sqrt{a}+\sqrt{b}}$

Answer 1

Answer: (D)

$\frac{m+n}{2}=\sqrt{ab}=\frac{ma+nb}{m+n}$
Taking $\sqrt{ab}=\frac{ma+mb}{m+n}$
$\Rightarrow \left(m+n\right)\sqrt{ab}=ma+nb$
$\Rightarrow m\left(\sqrt{ab}-a\right)=n\left(b-\sqrt{ab}\right)$
$\Rightarrow m\left(\sqrt{b}-\sqrt{a}\right)\sqrt{a}=n\left(\sqrt{b}-\sqrt{a}\right)\sqrt{b}$
$\Rightarrow \frac{m}{\sqrt{b}}=\frac{n}{\sqrt{a}}=k$ (say)
Further $\frac{m+n}{2}=\sqrt{ab}$
From $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{k}{2}\left(\sqrt{b}+\sqrt{a}\right)=\sqrt{ab}$
$\Rightarrow k=\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$
$\Rightarrow m=k\sqrt{b}=\frac{2b\sqrt{a}}{\sqrt{a}+\sqrt{b}}$.