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Find the value of $n$ so that $\large\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between $a$ and $b$.

$\begin{array}{1 1} \large\frac{1}{2} \\ \large\frac{-1}{2} \\ 1\\ -1\end{array} $

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  • G.M. $'g'$ between $a$ and $b$ is $\sqrt {ab}$
  • $(x+y)^2=x^2+y^2+2xy$
  • $\large\frac{x^m}{y^m}=\big(\frac{x}{y}\big)^m$
Given that $\large\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the G.M. between $a$ and $b$
We know that the G.M. between $a$ and $b$ =$\sqrt {ab}$
$\Rightarrow\: \large\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$$=\sqrt {ab}$
Squaring on both the sides we get
Step 2
Substituting the values in (i)
Taking $(a-b)$ common we get
We know that if $xy=0$ then either $x=0$ or $y=0$.
$\Rightarrow\:(a-b)=0$ or $(a^{2n+1}-b^{2n+1})=0$
But since $a\neq b$, $a^{2n+1}-b^{2n+1}=0$
We know that for any non zero number $x$, $x^0=1$
$\Rightarrow\:2n+1=0$ or $n=-\large\frac{1}{2}$
answered Mar 4, 2014 by rvidyagovindarajan_1

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