Question

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

H.M between $a$ and $b$

• A. $n=-1$
• B. $n=3$
• C. $n=-1/2$
• D. none of these

Answer 1

The correct option is A $n=-1$

$\because$ Harmonic Mean(H.M) between $a$ and $b$ is $\frac{2ab}{a+b}$

We have,

$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\frac{2ab}{a+b}$
$\Rightarrow (a^{n+1}+b^{n+1})(a+b)=2ab(a^n+b^n)$
$\Rightarrow a^{n+2}+a^{n+1}b+b^{n+1}a+b^{n+2}=2a^{n+1}b+2a{b}^{n+1}$
$\Rightarrow a^{n+2}-a^{n+1}b-a{b}^{n+1}+b^{n+2}=0$
$\Rightarrow a^{n+1}(a-b)-b^{n+1}(a-b)=0$
$\Rightarrow (a-b)(a^{n+1}-b^{n+1})=0$
$\because a-b\ne 0$
$\therefore a^{n+1}-b^{n+1}=0$
$\Rightarrow a^{n+1}=b^{n+1}$
$\Rightarrow {\left(\frac{a}{b}\right)}^{n+1}=1={\left(\frac{a}{b}\right)}^0$
Hence, $n+1=0$

$\therefore n=-1$