Question

If $\frac{\left(a^{n+1}+b^{n+1}\right)}{\left(a^n+b^n\right)}$ is the H.M of $a$ and $b$ then $n=\left(a,b\in R^{+},a\ne b\right)$.

• A. $0$
• B. $-1$
• C. $-\frac{1}{2}$
• D. $-2$

Answer 1

The correct option is B

$-1$

Finding the value of $n$:

Given $\frac{\left(a^{n+1}+b^{n+1}\right)}{\left(a^n+b^n\right)}$

We know HM of $a$ and $b$ is $\frac{2ab}{\left(a+b\right)}$

$\Rightarrow$ $\frac{\left(a^{n+1}+b^{n+1}\right)}{\left(a^n+b^n\right)}=\frac{2ab}{\left(a+b\right)}$

$\Rightarrow$$a^{n+2}+b^{n+2}+b{a}^{n+1}+a{b}^{n+1}=2b{a}^{n+1}+2a{b}^{n+1}$

$\Rightarrow$$a^{n+2}-b{a}^{n+1}+b^{n+2}-a{b}^{n+1}=0$

$\Rightarrow$ $a^{n+1}\left(a-b\right)+\left(-b^{n+1}\right)\left(a-b\right)=0$

$\Rightarrow$ $\left(a^{n+1}-b^{n+1}\right)\left(a-b\right)=0$

$\Rightarrow$ $a^{n+1}=b^{n+1}\left(\because a\ne b\right)$

$\Rightarrow$ $\frac{a^{n+1}}{b^{n+1}}=1$

$\Rightarrow$ $\frac{a^{n+1}}{b^{n+1}}={\left(\frac{a}{b}\right)}^0\left(\because n^0=1\right)$

$\Rightarrow$ $n+1=0$

$\Rightarrow$ $\mathit{n}\mathbf{=}\mathbf{-}\mathbf{1}$

Hence, correct answer is option $\left(B\right)$.