Question

If $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the A.M. between $a$ and $b,$ then find the value of $n.$

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $1$

A.M of $a$ and $b$ $=\frac{a+b}{2}$

According to the given condition,

$\frac{a+b}{2}=\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$

$(a+b)(a^{n-1}+b^{n-1})=2(a^n+b^n)$

$a^n+a{b}^{n-1}+b{a}^{n-1}+b^n=2a^n+2b^n$

$a{b}^{n-1}+a^{n-1}b=a^n+b^n$

$a{b}^{n-1}-b^n=a^n-a^{n-1}b$

$b^{n-1}(a-b)=a^{n-1}(a-b)$

$b^{n-1}=a^{n-1}$

$(\frac{a}{b}{)}^{n-1}=1$

$⟹n-1=0$

$n=1$