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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 $a\;\xi\;b$

$(a)\;n=\frac{-1}{3}\qquad(b)\;n=3\qquad(c)\;n=2\qquad(d)\;n=\frac{-1}{2}$

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Answer : (d) $n=\frac{-1}{2}$
$Explanation :\; Since\quad\;\frac{a^n+1+b^n+1}{a^n+b^n}\;is\;the\;GM\;between\;a\;\xi\;b$
$\frac{a^n+1+b^n+1}{a^n+b^n}\;=\sqrt{ab}$
$a^{n+1}+b^{n+1}=a^{n+1/2}\;b^{1/2}+a^{1/2}\;b^{n+1/2}$
$a^{n+1}-a^{n+1/2}\;b^{1/2}=a^{1/2}\;b^{n+1/2}-b^ {(n+1)}$
$a^{(n+1/2)} \;[a^{1/2}-b^{1/2}]=b^{(n+1/2)}\;[a^{1/2}-b^{1/2}]$
$Since\quad\;a\;\neq\;b\;,divide\;both\;sides\;by\;[a^{1/2}-b^{1/2}]$
$a^{n+1/2}=b^{n+1/2}\quad\;(a/b)^{n+1/2}$
$(a/b)^{n+1/2}=(a/b)^{0}$
$n+1/2=0\quad\;n=\frac{-1}{2}.$
answered Jan 3, 2014 by yamini.v
 

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