Question

If the geometric mean between $a$ and $b$ is $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$, then $n=$

• A. $0$
• B. $1/2$
• C. $-1/2$
• D. $1/4$

Answer 1

Answer: (C)

$-1/2$

$\left[\right(a{)}^{n+1}+\left(b{)}^{n+1}\right]=\left(ab{)}^{1/2}\right((a{)}^n+(b{)}^n)$
$(a{)}^{n+1}+(b{)}^{n+1}=(a{)}^{n+1/2}{b}^{1/2}+a^{(}1/2)(b{)}^{n+1/2}$
$(a{)}^{n+1}+(b{)}^{n+1}-(a{)}^{n+1/2}{b}^{1/2}-a^{(}1/2)(b{)}^{n+1/2}=0$
$\left(\right(a{)}^{n+1/2}-\left(b{)}^{n+1/2}\right)(a^{1/2}-b^{1/2})=0$
$(a{)}^{n+1/2}=(b{)}^{n+1/2}$ this is only possible when $n+1/2=0,n=-1/2$
Option C