Question

If the arithmetic mean of a and 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

Answer: (C)

$1$

We know that arithmetic mean between a & b is

$AM=\frac{a+b}{2}$

It is given that AM between a & b $=\frac{a^n+b^n}{a^{n-1}+b^{n-1}}$

So,

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

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

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

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

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

$a^n+b^n-a{b}^{n-1}-a^{n-1}b=0$

$a.a^{n-1}-a^{n-1}b+b.b^{n-1}-a{b}^{n-1}=0$

$a^{n-1}(a-b)-b^{n-1}(a-b)=0$

$(a^{n-1}-b^{n-1})(a-b)=0$

$a-b=0$

$a=b$ [not possible as $a\ne b$]

$\therefore a^{n-1}-b^{n-1}=0$

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

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

${\left(\frac{a}{b}\right)}^{n-1}={\left(\frac{a}{b}\right)}^0$ $[As{\left(\frac{a}{b}\right)}^0=1]$

Comparing powers

$n-1=0$

$n=1$