If a n + b n a n - 1 + b n - 1 is a G.M. between a and b , then the value of n is

Question

If $\frac{a^{n} + b^{n}}{a^{n - 1} + b^{n - 1}}$ is a G.M. between $a$ and $b$ , then the value of $n$ is

Smart Achievers · Accepted Answer

(4)

$Given, \frac{a^{n} + b^{n}}{a^{n - 1} + b^{n - 1}} = \sqrt{a b}$

$\Rightarrow$ $\frac{a^{n} + b^{n}}{\sqrt{a b}} = a^{n - 1} + b^{n - 1}$

$\Rightarrow$ $\frac{a^{n - \frac{1}{2}}}{\sqrt{b}} + \frac{b^{n - \frac{1}{2}}}{\sqrt{a}} - a^{n - 1} - b^{n - 1} = 0$

$\Rightarrow$ $a^{n - \frac{1}{2}} (b^{- \frac{1}{2}} - a^{- \frac{1}{2}}) + b^{n - \frac{1}{2}} (a^{- \frac{1}{2}} - b^{- \frac{1}{2}}) = 0$

$\Rightarrow$ $(b^{- \frac{1}{2}} - a^{- \frac{1}{2}}) (a^{n - \frac{1}{2}} - b^{n - \frac{1}{2}}) = 0$

$\Rightarrow$ $b^{- \frac{1}{2}} - a^{- \frac{1}{2}} = 0 \Rightarrow \sqrt{a} = \sqrt{b}$

or $a^{n - \frac{1}{2}} - b^{n - \frac{1}{2}} = 0$

$\Rightarrow$ ${(\frac{a}{b})}^{n - \frac{1}{2}} = 1 = {(\frac{a}{b})}^{0} \Rightarrow n = \frac{1}{2}$

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