If $\frac{(a^{n+1}+b^{n+1})}{(a^n+b^n)}$ be the A.M of $a$ and $b$, then $n=$

The correct option is A

$1$

Explanation for the correct option:

Step 1. Find the value of $n$:

Arithmetic mean of $a$ and $b$, AM $=\frac{(a+b)}{2}$

Given that AM $=\frac{(a^n+b^n)}{(a^{n-1}+b^{n-1})}$

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

Step 2. By Cross multiplying, we get

$2(a^n+b^n)=(a+b)(a^{n-1}+b^{n-1})$

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

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

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

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

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

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

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

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

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

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

Step 3. By Comparing powers, we get

$n-1=0$

$\mathbf{\therefore }\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘A’ is Correct.