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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Given that $a\;\xi\;b$ are not equal , find the value of n so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the GM of $a\;\xi\;b$


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Answer : (d) $n=\frac{-1}{2}$
$Explanation :\; Since\quad\;\frac{a^n+1+b^n+1}{a^n+b^n}\;is\;the\;GM\;between\;a\;\xi\;b$
$a^{n+1}-a^{n+1/2}\;b^{1/2}=a^{1/2}\;b^{n+1/2}-b^ {(n+1)}$
$a^{(n+1/2)} \;[a^{1/2}-b^{1/2}]=b^{(n+1/2)}\;[a^{1/2}-b^{1/2}]$
answered Jan 3, 2014 by yamini.v

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