Question

For what value of $n,$ $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is $(a\ne b)$

G.M between $a$ and $b$.

• A. $n=3$
• B. $n=-1$
• C. $n=-\frac{1}{2}$
• D. none of these

Answer 1

The correct option is C $n=-\frac{1}{2}$

$\because G.M$ between $a$ and $b$ is $\sqrt{ab}$
we have,
$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab}$
$\Rightarrow a^{n+1}+b^{n+1}=\sqrt{ab}(a^n+b^n)$
$\Rightarrow a^{n+1}+b^{n+1}=a^{n+\frac{1}{2}}.b^{\frac{1}{2}}+a^{\frac{1}{2}}.b^{n+\frac{1}{2}}$
$\Rightarrow a^{n+\frac{1}{2}}(a^{\frac{1}{2}}-b^{\frac{1}{2}})-b^{n+\frac{1}{2}}(a^{\frac{1}{2}}-b^{\frac{1}{2}})=0$
$\Rightarrow (a^{\frac{1}{2}}-b^{\frac{1}{2}})(a^{n+\frac{1}{2}}-b^{n+\frac{1}{2}})=0$
$\because a^{\frac{1}{2}}-b^{\frac{1}{2}}\ne 0$
$\therefore a^{n+\frac{1}{2}}-b^{n+\frac{1}{2}}=0$
$\Rightarrow {\left(\frac{a}{b}\right)}^{n+\frac{1}{2}}=1={\left(\frac{a}{b}\right)}^0$
$\therefore n+\frac{1}{2}=0$
Hence, $n=-\frac{1}{2}$