Question

For what value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is the geometric mean of $a$ and $b$

Answer 1

We know that the geometric mean between $a$ and $b$ is $=\sqrt{ab}$

G.M between $a$ and $b=\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$

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

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

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

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

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

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

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

$\Rightarrow {\left(\frac{a}{b}\right)}^{n+\frac{1}{2}}={\left(\frac{a}{b}\right)}^0$

Comparing the powers, we get

$n+\frac{1}{2}=0$

$\therefore n=\frac{-1}{2}$