Question

If the arithmetic mean of $a\&b$is $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$, then the value of $n$ is:

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

The correct option is C

$1$

Explanation for the correct option.

Step 1: Simplify the expression of mean
It is given that $\mathrm{AM}$ between $\mathrm{a}\&\mathrm{b}=\frac{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n}-1}+{\mathrm{b}}^{\mathrm{n}-1}}$.

We know that the arithmetic mean between $a\&b$ is: $\mathrm{AM}=\frac{\mathrm{a}+\mathrm{b}}{2}$

$\begin{array}{rcl}& \Rightarrow & \frac{a^n+b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}\\ & \Rightarrow & 2a^n+2b^n=a\left(a^{n-1}+b^{n-1}\right)+b\left(a^{n-1}+b^{n-1}\right)\\ & \Rightarrow & 2a^n+2b^n=a{a}^{n-1}+a{b}^{n-1}+b{a}^{n-1}+b{b}^{n-1}\\ & \Rightarrow & 2a^n+2b^n=a^n+a{b}^{n-1}+b{a}^{n-1}+b^n\\ & \Rightarrow & a^n+b^n-a{b}^{n-1}-a^{n-1}b=0\\ & \Rightarrow & a·a^{n-1}-a^{n-1}b+b·b^{n-1}-a{b}^{n-1}=0\\ \Rightarrow a^{n-1}(a-b)-b^{n-1}(a-b)& =& 0\\ & \Rightarrow & \left(a^{n-1}-b^{n-1}\right)(a-b)=0\\ & \Rightarrow & a-b=0\\ & \Rightarrow & a=b[not\text{possible as a}\equiv b]\end{array}$

Step 2: Find the value of $n$

$\begin{array}{rcl}{a}^{n-1}-b^{n-1}& =& 0\\ & \Rightarrow & a^{n-1}=b^{n-1}\\ & \Rightarrow & \frac{a^{n-1}}{b^{n-1}}=1\\ & \Rightarrow & {\left(\frac{a}{b}\right)}^{n-1}={\left(\frac{a}{b}\right)}^0\\ & \Rightarrow & n-1=0\\ & \Rightarrow & n=1\end{array}$

Hence, option C is correct.