MCQ
Let $G$ be the geometric mean of two positive numbers $a$ and $b,$ and $M$ be the arithmetic mean of $\frac {1}{a}$ and $\frac {1}{b}$. If $\frac {1}{M}\,:\,G$ is $4:5,$ then $a:b$ can be
- ✓$1:4$
- B$1:2$
- C$2:3$
- D$3:4$
$M = \frac{{\frac{1}{a} + \frac{1}{b}}}{2}$
$M = \frac{{a + b}}{{2ab}}$
Given that $\frac{1}{M}:G = 4:5$
$\frac{{2ab}}{{\left( {a + b} \right)\sqrt {ab} }} = \frac{4}{5}$
$ \Rightarrow \frac{{a + b}}{{2\sqrt {ab} }} = \frac{5}{4}$
$ \Rightarrow \frac{{a + b + 2\sqrt {ab} }}{{a + b - 2\sqrt {ab} }} = \frac{{5 + 4}}{{5 - 4}}$
{Using Componendo & Dividendo}
$ \Rightarrow \frac{{{{\left( {\sqrt a } \right)}^2} + {{\left( {\sqrt b } \right)}^2} + 2\sqrt {ab} }}{{{{\left( {\sqrt a } \right)}^2} + {{\left( {\sqrt b } \right)}^2} - 2\sqrt {ab} }} = \frac{9}{1}$
$ \Rightarrow {\left( {\frac{{\sqrt b + \sqrt a }}{{\sqrt b - \sqrt a }}} \right)^2} = \frac{9}{1} \Rightarrow \frac{{\sqrt b + \sqrt a }}{{\sqrt b - \sqrt a }} = \frac{3}{1}$
$ \Rightarrow \frac{{\sqrt b + \sqrt a + \sqrt b - \sqrt a }}{{\sqrt b + \sqrt a - \sqrt b + \sqrt a }} = \frac{{3 + 1}}{{3 - 1}}$
{Using Componendo & Dividendo}
$\sqrt {\frac{b}{a}} = \frac{4}{2} = 2$
$\frac{b}{a} = \frac{4}{1}$
$\frac{a}{b} = \frac{1}{4} \Rightarrow a:b = 1:4$
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