Question

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

Answer 1

Given:- $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is arithmetic mean of a and b.
Since arithmetic mean between a and b is $\frac{a+b}{2}$
$$\begin{gathered} \Rightarrow \frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\frac{a+b}{2} \\ 2\left(a^{n+1}+b^{n+1}\right)=\left(a+b\right)\left(a^n+b^n\right) \\ \mathrm{i}.\mathrm{e}2a^{n+1}+2b^{n+1}=a^{n+1}+a{b}^n+b{a}^n+b^{n+1} \\ \mathrm{i}.\mathrm{e}{a}^{n+1}+b^{n+1}=a{b}^n+b{a}^n \\ \mathrm{i}.\mathrm{e}{a}^{n+1}-a^{n}b=a{b}^n-b^{n+1} \\ \mathrm{i}.\mathrm{e}{a}^n\left(a-b\right)=b^n\left(a-b\right) \\ \mathrm{i}.\mathrm{e}{a}^n=b^n\left(\mathrm{Since}a-b\ne 0aandb\mathrm{are}\mathrm{distinct}\right) \\ \mathrm{i}.\mathrm{e}{\left(\frac{a}{b}\right)}^n=1 \\ \mathrm{i}.\mathrm{e}n=0 \end{gathered}$$