If the ratio of A.M. between two positive rea | vedclass

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(C) Given that the ratio of $A.M.$ to $H.M.$ is $m:n$,we have $\frac{(a+b)/2}{2ab/(a+b)} = \frac{m}{n}$.This simplifies to $\frac{(a+b)^2}{4ab} = \fra

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$\frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}$

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(C) Given that the ratio of $A.M.$ to $H.M.$ is $m:n$,we have $\frac{(a+b)/2}{2ab/(a+b)} = \frac{m}{n}$.This simplifies to $\frac{(a+b)^2}{4ab} = \frac{m}{n}$,which implies $\frac{(a+b)^2}{ab} = \frac{4m}{n}$.Using the property of componendo and dividendo on $\frac{(a+b)^2}{4ab} = \frac{m}{n}$,we get $\frac{(a+b)^2 + 4ab}{(a+b)^2 - 4ab} = \frac{m+n}{m-n}$.This simplifies to $\frac{(a+b)^2}{(a-b)^2} = \frac{m+n}{m-n}$ is incorrect. Let us re-evaluate: $\frac{(a+b)^2}{4ab} = \frac{m}{n} \Rightarrow \frac{4ab}{(a+b)^2} = \frac{n}{m}$.Applying componendo and dividendo: $\frac{(a+b)^2 + 4ab}{(a+b)^2 - 4ab} = \frac{m+n}{m-n} \Rightarrow \frac{a^2+b^2+6ab}{a^2+b^2-2ab} = \frac{m+n}{m-n}$ (not helpful).Correct approach: $\frac{(a+b)^2}{4ab} = \frac{m}{n} \Rightarrow \frac{(a+b)^2}{ab} = \frac{4m}{n}$.We know $(a+b)^2 - (a-b)^2 = 4ab$,so $\frac{(a+b)^2}{(a-b)^2} = \frac{m}{m-n}$.Taking square root,$\frac{a+b}{a-b} = \frac{\sqrt{m}}{\sqrt{m-n}}$.Applying componendo and dividendo again: $\frac{(a+b)+(a-b)}{(a+b)-(a-b)} = \frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}$.Thus,$\frac{2a}{2b} = \frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}$,so $\frac{a}{b} = \frac{\sqrt{m} + \sqrt{m-n}}{\sqrt{m} - \sqrt{m-n}}$.

If the ratio of $A.M.$ between two positive real numbers $a$ and $b$ to their $H.M.$ is $m:n$,then $a:b$ is

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