Let $\mathrm{G}$ be the geometric mean of two positive numbers $\mathrm{a}$ and $\mathrm{b}$, and $\mathrm{M}$ be the arithmetic mean of $\frac{1}{\mathrm{a}}$ and $\frac{1}{\mathrm{~b}}$. If $\frac{1}{\mathrm{M}}: \mathrm{G}$ is $4: 5$, then $\mathrm{a}: \mathrm{b}$ can be:

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Aicharya · Accepted Answer

$\mathrm{G}=\sqrt{a b}$ $ \begin{aligned} &\mathrm{M}=\frac{\frac{1}{a}+\frac{1}{b}}{2} \ &\mathrm{M}=\frac{a+b}{2 a b} \end{aligned} $ Given that $\frac{1}{\mathrm{M}}: \mathrm{G}=4: 5$ $ \begin{aligned} & \frac{2 a b}{(a+b) \sqrt{a b}}=\frac{4}{5} \ \Rightarrow & \frac{a+b}{2 \sqrt{a b}}=\frac{5}{4} \ \Rightarrow & \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{5+4}{5-4} \end{aligned} $ \{Using Componendo \& Dividendo $ \begin{aligned} &\Rightarrow \frac{(\sqrt{a})^2+(\sqrt{b})^2+2 \sqrt{a b}}{(\sqrt{a})^2+(\sqrt{b})^2-2 \sqrt{a b}}=\frac{9}{1} \ &\Rightarrow\left(\frac{\sqrt{b}+\sqrt{a})^2}{\sqrt{b}-\sqrt{a}}ight)^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} \end{aligned} $ \{Using Componendo \& Dividendo $ \sqrt{\frac{b}{a}}=\frac{4}{2}=2 $ $ \begin{aligned} \frac{b}{a} &=\frac{4}{1} \ \frac{a}{b} &=\frac{1}{4} \Rightarrow a: b=1: 4 \end{aligned} $

Let $\mathrm{G}$ be the geometric mean of two positive numbers $\mathrm{a}$ and $\mathrm{b}$, and…

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