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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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If $\large\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b$, then find the value of $n$

$\begin{array}{1 1}0 \\ 1\\ 2\\ 3 \end{array} $

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  • The $A.M.$ between $a\:\;and\:\:b$ is $\large\frac{a+b}{2}$
Given that $\large\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b$.
We know that the $A.M.$ between $a\:\;and\:\:b$ is $\large\frac{a+b}{2}$
$\Rightarrow\:\large\frac{a+b}{2}=\large\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$
$\Rightarrow\:(a+b)(a^{n-1}+b^{n-1})=2(a^n+b^n)$
$\Rightarrow\:a^n+ab^{n-1}+ba^{n-1}+b^n=2a^n+2b^n$
$\Rightarrow\:a^n+b^n-ab^{n-1}-ba^{n-1}=0$
$\Rightarrow\:(a^n-ba^{n-1})-(ab^{n-1}-b^n)=0$
$\Rightarrow\:a^{n-1}(a-b)-b^{n-1}(a-b)=0$
$\Rightarrow\:(a-b)(a^{n-1}-b^{n-1})=0$
$\Rightarrow\:a-b=0\:\:or\:\:a^{n-1}-b^{n-1}=0$
But since $a\neq b$, $a^{n-1}=b^{n-1}$
$\Rightarrow\:\large\frac{a^{n-1}}{b^{n-1}}$$=1$
$\Rightarrow\:\bigg(\large\frac{a}{b}\bigg)^{n-1}$$=1$
$\Rightarrow\:n-1=0$ or $n=1$
answered Feb 24, 2014 by rvidyagovindarajan_1
 

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