Question

The value of for which $\dfrac{{{x}^{n+1}}+{{y}^{n+1}}}{{{x}^{n}}+{{y}^{n}}}$ is the geometric mean of $x$ and $y$ is
A. $n=-\dfrac{1}{2}$
B. $n=\dfrac{1}{2}$
C. $n=1$
D. $n=-1$

Answer 1

Hint: We first find the geometric mean for two values $x$ and $y$ as ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}$. We reform the term with respect to the given expression of $\dfrac{{{x}^{n+1}}+{{y}^{n+1}}}{{{x}^{n}}+{{y}^{n}}}$. We equate it with the value of $n$ and find the solution.

Complete step-by-step answer:
We have to find the value of $n$ for which $\dfrac{{{x}^{n+1}}+{{y}^{n+1}}}{{{x}^{n}}+{{y}^{n}}}$ is the geometric mean of $x$ and $y$.
We know that for two values $x$ and $y$, the geometric mean will be $\sqrt{xy}={{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}$.
Now we have to somehow change ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}$ to equate with the form of $\dfrac{{{x}^{n+1}}+{{y}^{n+1}}}{{{x}^{n}}+{{y}^{n}}}$.
We can write ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}={{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}\times \dfrac{{{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}}}{{{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}}}$.
Now we put ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}$ in the denominator.
So, \[{{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}=\dfrac{{{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}}}{\dfrac{{{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}}}{{{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}}}\].
We now simplify \[\dfrac{{{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}}}{{{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}}\]. We get
\[\dfrac{{{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}}}{{{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}}={{x}^{-\dfrac{1}{2}}}+{{y}^{-\dfrac{1}{2}}}\] .
So, we can write ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}=\dfrac{{{x}^{\dfrac{1}{2}}}+{{y}^{\dfrac{1}{2}}}}{{{x}^{-\dfrac{1}{2}}}+{{y}^{-\dfrac{1}{2}}}}=\dfrac{{{x}^{-\dfrac{1}{2}+1}}+{{y}^{-\dfrac{1}{2}+1}}}{{{x}^{-\dfrac{1}{2}}}+{{y}^{-\dfrac{1}{2}}}}$.
Equating $\dfrac{{{x}^{-\dfrac{1}{2}+1}}+{{y}^{-\dfrac{1}{2}+1}}}{{{x}^{-\dfrac{1}{2}}}+{{y}^{-\dfrac{1}{2}}}}$ with $\dfrac{{{x}^{n+1}}+{{y}^{n+1}}}{{{x}^{n}}+{{y}^{n}}}$, we get $n=-\dfrac{1}{2}$. The correct option is A.
So, the correct answer is “Option A”.

Note: We also can use the option check to simplify the problem. We use the options one by one and simplify if the form becomes the geometric mean two values $x$ and $y$ as ${{x}^{\dfrac{1}{2}}}{{y}^{\dfrac{1}{2}}}$. We can also use the same process to find the arithmetic mean and the harmonic mean.