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

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(A) $G = \sqrt{ab}$$M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a + b}{2ab}$Given that $\frac{1}{M} : G = 4 : 5$,we have $\frac{2ab}{(a + b)\sqrt{

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$1:4$

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(A) $G = \sqrt{ab}$$M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a + b}{2ab}$Given that $\frac{1}{M} : G = 4 : 5$,we have $\frac{2ab}{(a + b)\sqrt{ab}} = \frac{4}{5}$$\Rightarrow \frac{a + b}{2\sqrt{ab}} = \frac{5}{4}$Using Componendo and Dividendo:$\frac{a + b + 2\sqrt{ab}}{a + b - 2\sqrt{ab}} = \frac{5 + 4}{5 - 4}$$\Rightarrow \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = 9$$\Rightarrow \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = 3$ (assuming $a > b$ for positive ratio)$\Rightarrow \sqrt{a} + \sqrt{b} = 3\sqrt{a} - 3\sqrt{b}$$\Rightarrow 4\sqrt{b} = 2\sqrt{a}$ $\Rightarrow \sqrt{\frac{a}{b}} = 2$ $\Rightarrow \frac{a}{b} = 4$Since the ratio $a:b$ can be $1:4$ or $4:1$,and $1:4$ is given in the options,the correct answer is $1:4$.

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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