Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b .$
Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b .$
$M$. of $a$ and $b$ is $\sqrt{a b}$
By the given condition: $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{a b}$
Squaring both sides, we obtain
$\frac{\left(a^{n+1}+b^{n+1}\right)^{2}}{\left(a^{n}+b^{n}\right)^{2}}=a b$
$\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n+2}=(a b)\left(a^{2 n}+2 a^{n} b^{n}+b^{2 n}\right)$
$\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n+2}=a^{2 n+1} b+2 a^{n+1} b^{n+1}+a b^{2 n+1}$
$\Rightarrow a^{2 n+2}+b^{2 n+2}=a^{2 n+1} b+a b^{2 n+1}$
$\Rightarrow a^{2 n+2}-a^{2 n+1} b=a b^{2 n+1}-b^{2 n+2}$
$\Rightarrow a^{2 n+1}(a-b)=b^{2 n+1}(a-b)$
$\Rightarrow\left(\frac{a}{b}\right)^{2 n+1}=1=\left(\frac{a}{b}\right)^{0}$
$\Rightarrow 2 n+1=0$
$\Rightarrow n=\frac{-1}{2}$
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