Question

If the H.M. of two distinct numbers $a$ and $b$ is $\frac{a^{n-1}+b^{n-1}}{a^{n-2}+b^{n-2}}$, then the value of $n$ is

Answer 1

The H.M. of $a$ and $b$
$=\frac{2ab}{a+b}$

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