Question

If the A.M of $a\text{and}b$ then $\frac{\left(a^{n+1}+b^{n+1}\right)}{a^n+b^n}$ then $n=$

• A. $2$
• B. $3$
• C. 1
• D. $0$

Answer 1

The correct option is C

1

Explanation for the correct option

Step 1: Information required for the solution

The arithmetic mean (A.M) between two numbers $a\text{and}b$ is calculated as

$\mathbf{\text{A.M}}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots \left(1\right)$

Here, it is given that the A.M between such two numbers is

$\begin{array}{rcl}\text{A.M}& =& \frac{\left(a^{n+1}+b^{n+1}\right)}{a^n+b^n}\\ & =& \frac{a^{n}a+b^{n}b}{a^n+b^n}\\ & =& \frac{a^n+b^n}{a^{n-1}+b^{n-1}}\end{array}$

From this, we get the second equation as $\text{A.M}=\begin{array}{rcl}& & \frac{a^n+b^n}{a^{n-1}+b^{n-1}}\end{array}$

Step 2: Determination of the value of $\mathit{n}$

From $\left(1\right)$ and $\left(2\right)$,

$\begin{array}{rcl}\therefore \frac{a+b}{2}& =& \frac{a^n+b^n}{a^{n-1}+b^{n-1}}\\ & \Rightarrow & \left(a+b\right)\left(a^{n-1}+b^{n-1}\right)=2\left(a^n+b^n\right)\\ & \Rightarrow & a^n+a{b}^{n-1}+b{a}^{n-1}+b^n=2a^n+2b^n\\ & \Rightarrow & a{b}^{n-1}+b{a}^{n-1}=a^n+b^n\end{array}$

Take all the expressions on the same side,

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

From this, can see that

$\begin{array}{rcl}\therefore a^{n-1}-b^{n-1}& =& 0\\ & \Rightarrow & a^{n-1}=b^{n-1}\end{array}$

Divide bot the sides by $\begin{array}{rcl}& & b^{n-1}\end{array}$,

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

We can conclude that,

$\begin{array}{rcl}\therefore n-1& =& 0\\ & \Rightarrow & n=1\end{array}$

Hence, the correct answer is an option (C)