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

Answer 1

The correct option is A

$-\frac{1}{2}$

Explanation for the correct answer:

Compute the required value:

We know that the geometric mean of $x$ and $y$ is $\sqrt{xy}$

According to the given condition,

$\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=\sqrt{xy}$

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

$\Rightarrow$ $x.x^n+y.y^n=x^{\frac{1}{2}}{y}^{\frac{1}{2}}\left(x^n+y^n\right)$

$\Rightarrow$ $x.x^n+y.y^n=x^{n+\frac{1}{2}}{y}^{\frac{1}{2}}+x^{\frac{1}{2}}{y}^{n+\frac{1}{2}}$

$\Rightarrow$ $x.x^n+y.y^n=x^n{x}^{\frac{1}{2}}{y}^{\frac{1}{2}}+x^{\frac{1}{2}}{y}^n{y}^{\frac{1}{2}}$

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

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

$\Rightarrow$ $\frac{x^{n+{\displaystyle \frac{1}{2}}}}{y^{n+{\displaystyle \frac{1}{2}}}}=1$

$\Rightarrow$ ${\left(\frac{x}{y}\right)}^{n+\frac{1}{2}}=1$ $\left[\because a^0=1\right]$

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

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

Hence, $\frac{x^{n+1}+y^{n+1}}{x^n+y^n}$ is the geometric mean of $x$ and $y$ when $n$ is $-\frac{1}{2}$.

Hence, option $\left(A\right)$ is the correct answer.