Question

If $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the A.M. between $a$ and $b$, then find the value of $n$.

Answer 1

The A.M. between $a$ and $b$ is given by.
$\text{A.M}=\frac{a+b}{2}$
Then according to the question.
$\frac{a+b}{2}=\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$
$\Rightarrow (a+b)(a^{n-1}+b^{n-1})=2(a^n+b^n)$
$\Rightarrow a^n+a{b}^{n-1}+b{a}^{n-1}+b^n=2a^n+2b^n$
$\Rightarrow a{b}^{n-1}+a^{n-1}b=a^n+b^n$
$\Rightarrow a{b}^{n-1}-b^n=a^n-a^{n-1}b$
$\Rightarrow b^{n-1}(a-b)=a^{n-1}(a-b)$
$\Rightarrow b^{n-1}=a^{n-1}\{\because a\ne b\}$
$\Rightarrow {\left(\frac{a}{b}\right)}^{n-1}=1={\left(\frac{a}{b}\right)}^0$
By comparing powers, we get
$n-1=0$
$\therefore n=1$

Finla answer: Hence, the value of $n$ is $1$