If the ratio of the harmonic mean to the geom | vedclass

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(C) Let the two positive numbers be $a$ and $b$. The geometric mean $(GM)$ is $\sqrt{ab}$ and the harmonic mean $(HM)$ is $\frac{2ab}{a+b}$. Given the

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the two positive numbers be $a$ and $b$. The geometric mean $(GM)$ is $\sqrt{ab}$ and the harmonic mean $(HM)$ is $\frac{2ab}{a+b}$. Given the ratio $\frac{HM}{GM} = \frac{4}{5}$. Substituting the formulas: $\frac{\frac{2ab}{a+b}}{\sqrt{ab}} = \frac{4}{5}$. This simplifies to $\frac{2\sqrt{ab}}{a+b} = \frac{4}{5}$,which implies $\frac{\sqrt{ab}}{a+b} = \frac{2}{5}$. Squaring both sides: $\frac{ab}{(a+b)^2} = \frac{4}{25}$. Let $x = \frac{a}{b}$. Then $\frac{b^2 x}{b^2(x+1)^2} = \frac{4}{25}$,so $\frac{x}{(x+1)^2} = \frac{4}{25}$. $25x = 4(x^2 + 2x + 1) \implies 4x^2 - 17x + 4 = 0$. Factoring the quadratic: $(4x - 1)(x - 4) = 0$. Since $a > b$,$x = \frac{a}{b} > 1$,so $x = 4$. Therefore,$a : b = 4 : 1$.

If the ratio of the harmonic mean to the geometric mean of two positive numbers $a$ and $b$ $(a > b)$ is $4 : 5$,then $a : b = \dots$ (in $: 1$)

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