Question

For what value of $n,$ $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is $(a\ne b)$

A.M between $a$ and $b$ is

• A. $n=-1$
• B. $n=-\frac{1}{2}$
• C. $n=3$
• D. none of these

Answer 1

The correct option is D none of these

$\because A.M$ between $a$ and $b$ is $\frac{a+b}{2}$
we have $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\frac{a+b}{2}$
$\Rightarrow 2a^{n+1}+2b^{n+1}=(a+b)(a^n+b^n)$
$\Rightarrow 2a^{n+1}+2b^{n+1}=a^{n+1}+a{b}^n+b{a}^n+b^{n+1}$
$\Rightarrow a^{n+1}-a{b}^n+b^{n+1}-b{a}^n=0$
$\Rightarrow a^n(a-b)-b^n(a-b)=0$
$\Rightarrow (a-b)(a^n-b^n)=0$
$\because a-b\ne 0$
$\therefore a^n-b^n=0$
$\Rightarrow {\left(\frac{a}{b}\right)}^n=1$
$\Rightarrow {\left(\frac{a}{b}\right)}^n={\left(\frac{a}{b}\right)}^0$
$\therefore n=0$