Question

Find the value of n so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between a and b.

Answer 1

Wew know that G.M between 'a' and 'b' is $\sqrt{ab}$

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

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

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

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

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

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

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

$\Rightarrow n+\frac{1}{2}=\Rightarrow n=-\frac{1}{2}.$