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

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(A) Given that $G$ is the geometric mean of $a$ and $b,$ so $G = \sqrt{ab}.$$M$ is the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b},$ so $M = \fr

Vedclass · Accepted Answer

$1:4$

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(A) Given that $G$ is the geometric mean of $a$ and $b,$ so $G = \sqrt{ab}.$$M$ is the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b},$ so $M = \frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{a+b}{2ab}.$Therefore,$\frac{1}{M} = \frac{2ab}{a+b}.$Given the ratio $\frac{1}{M}:G = 4:5,$ we have $\frac{2ab}{(a+b)\sqrt{ab}} = \frac{4}{5}.$Simplifying the expression,$\frac{2\sqrt{ab}}{a+b} = \frac{4}{5},$ which implies $\frac{a+b}{2\sqrt{ab}} = \frac{5}{4}.$Applying Componendo and Dividendo,$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{5+4}{5-4}.$This simplifies to $\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 9.$Taking the square root on both sides,$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = 3$ (assuming $a > b$) or $-3.$Solving $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = 3$ gives $\sqrt{a}+\sqrt{b} = 3\sqrt{a}-3\sqrt{b},$ so $4\sqrt{b} = 2\sqrt{a},$ which means $2\sqrt{b} = \sqrt{a}.$Squaring both sides,$4b = a,$ so $\frac{a}{b} = 4,$ which gives $a:b = 4:1.$If we take the ratio as $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = -3,$ we get $\sqrt{a}+\sqrt{b} = -3\sqrt{a}+3\sqrt{b},$ so $4\sqrt{a} = 2\sqrt{b},$ which means $2\sqrt{a} = \sqrt{b}.$Squaring both sides,$4a = b,$ so $\frac{a}{b} = \frac{1}{4},$ which gives $a:b = 1:4.$Thus,$a:b$ can be $1:4.$

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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