Question

If $\frac{(a^n+b^n)}{(a^{n-1}+b^{n-1})}$ is the G.M of $a$ and $b$ then $n=(a,b\in R+,a\ne b)$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $-2$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

Step 1. Find the value of $n$:

As we know,

G.M of $a$ and $b=\sqrt{\left(ab\right)}$

$\Rightarrow$ $\frac{(a^n+b^n)}{(a^{n-1}+b^{n-1})}=\sqrt{\left(ab\right)}$

$\Rightarrow$ $\frac{(a^n+b^n)}{\sqrt{\left(ab\right)}}=(a^{n-1}+b^{n-1})$

$\Rightarrow$ $a^{n-\frac{1}{2}}{b}^{-\frac{1}{2}}-a^{n-1}+b^{n-\frac{1}{2}}{a}^{-\frac{1}{2}}–b^{n-1}=0$

$\Rightarrow$$a^{n-\frac{1}{2}}\left(b^{-\frac{1}{2}}-a^{-\frac{1}{2}}\right)+b^{n-\frac{1}{2}}\left(a^{-\frac{1}{2}}-b^{-\frac{1}{2}}\right)=0$

$\Rightarrow$ $\left(b^{-\frac{1}{2}}-a^{-\frac{1}{2}}\right)\left(a^{n-\frac{1}{2}}-b^{n-\frac{1}{2}}\right)=0$

$\Rightarrow$ $\left(b^{-\frac{1}{2}}-a^{-\frac{1}{2}}\right)=0,\left(a^{n-\frac{1}{2}}-b^{n-\frac{1}{2}}\right)=0$

$\Rightarrow$$b^{-\frac{1}{2}}=a^{-\frac{1}{2}},(a^{n-\frac{1}{2}}=b^{n-\frac{1}{2}})$

$\Rightarrow$ $\sqrt{a}=\sqrt{b},{\left(\frac{a}{b}\right)}^{n-\frac{1}{2}}=1$

$\Rightarrow$ ${\left(\frac{a}{b}\right)}^{n-\frac{1}{2}}={\left(\frac{a}{b}\right)}^0$

Step 2. By Comparing powers, we get

$n-\frac{1}{2}=0$

$\mathbf{\therefore }\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}$

Hence, Option ‘C’ is Correct.