Question

The value of $n$ for which $\frac{x^{n+1}+y^{n+1}}{x^n+y^n}$ is the geometric mean of $x$ and $y$ is

• A. $n=-\frac{1}{2}$
• B. $n=\frac{1}{2}$
• C. $n=1$
• D. $n=-1$

Answer 1

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

Given, $\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=\sqrt{xy}$ .......... Geometric mean

$\Rightarrow x^{n+1}+y^{n+1}=\sqrt{xy}(x^n+y^n)$

$\Rightarrow x\cdot x^n-x^n\cdot x^{\frac{1}{2}}\cdot y^{\frac{1}{2}}=y^n\cdot x^{\frac{1}{2}}\cdot y^{\frac{1}{2}}-y\cdot y^n$

$\Rightarrow x^{n+\frac{1}{2}}\left(\begin{array}{c}{x}^{\frac{1}{2}}-y^{\frac{1}{2}}\end{array}\right)=y^{n+\frac{1}{2}}\left(\begin{array}{c}{x}^{\frac{1}{2}}-y^{\frac{1}{2}}\end{array}\right)$

$\Rightarrow {\left(\begin{array}{c}\frac{x}{y}\end{array}\right)}^{n+\frac{1}{2}}=1$

$\Rightarrow {\left(\begin{array}{c}\frac{x}{y}\end{array}\right)}^{n+\frac{1}{2}}={\left(\begin{array}{c}\frac{x}{y}\end{array}\right)}^0$

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