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}}$.