Let G be the geometric mean of two positive n | vedclass

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Question

$G=\sqrt {ab} $

$M = \frac{{\frac{1}{a} + \frac{1}{b}}}{2}$

$M = \frac{{a + b}}{{2ab}}$

Given that $\frac{1}{M}:G = 4:5$

$\frac{{2ab}

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

$G=\sqrt {ab} $

$M = \frac{{\frac{1}{a} + \frac{1}{b}}}{2}$

$M = \frac{{a + b}}{{2ab}}$

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}

$\sqrt {\frac{b}{a}}  = \frac{4}{2} = 2$

$\frac{b}{a} = \frac{4}{1}$

$\frac{a}{b} = \frac{1}{4} \Rightarrow a:b = 1:4$

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

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