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27. 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.

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M of a and b is $\sqrt{ab}.$

Given :

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

Squaring both sides ,

$\left ( \frac{a^{n+1}+ b ^{n+1}}{a^n+b^n} \right )^2=ab$

$\left (a^{n+1}+ b ^{n+1})^2=({a^n+b^n} \right )^2ab$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2}+2.a^{n+1}.b^{n+1})=({a^2^n+b^2^n+2.a^n.b^n} \right )ab$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2}+2.a^{n+1}.b^{n+1})=({a^{2n+1}.b+a.b^{2n+1}+2.a^{n+1}.b^{n+1}} \right )$

$\Rightarrow \left (a^{2n+2}+ b ^{2n+2})=({a^{2n+1}.b+a.b^{2n+1}} \right )$

$\Rightarrow a^{2n+2}-{a^{2n+1}.b=a.b^{2n+1}} \right )- b ^{2n+2}$

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

$\Rightarrow a^{2n+1}=b^{2n+1}$

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

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

$\Rightarrow 2n+1=0$

$\Rightarrow 2n=-1$

$\Rightarrow n=\frac{-1}{2}$

Posted by

seema garhwal

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