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

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Question

(c) $\frac{{\frac{{a + b}}{2}}}{{\frac{{2ab}}{{a + b}}}} = \frac{m}{n}$

$ \Rightarrow $ $\frac{{{{(a + b)}^2}}}{{4ab}} = \frac{m}{n}$

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Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $\frac{{\frac{{a + b}}{2}}}{{\frac{{2ab}}{{a + b}}}} = \frac{m}{n}$

$ \Rightarrow $ $\frac{{{{(a + b)}^2}}}{{4ab}} = \frac{m}{n}$

$ \Rightarrow $$\frac{{{{(a + b)}^2}}}{{2ab}} = \frac{{2m}}{n}$

Applying dividendo, we get $\frac{{{a^2} + {b^2}}}{{2ab}} = \frac{{2m - n}}{n}$

Applying componendo and dividendo,

We get $\frac{{{{(a + b)}^2}}}{{{{(a - b)}^2}}} = \frac{m}{{m - n}}$

$ \Rightarrow $$\frac{{a + b}}{{a - b}} = \frac{{\sqrt m }}{{\sqrt {m - n} }}$

Again, applying componendo and dividendo

We get $\frac{a}{b} = \frac{{\sqrt m + \sqrt {m - n} }}{{\sqrt m - \sqrt {m - n} }}$.

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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