Find the value of `n` so that `(a^(n+1)+b^(n+1))/(a^n+b^n)` may be the geometric mean between `a` and `b`

Options

A. `n=1/3`
B. `n=-1/2`
C. `n=-1/3`
D. `n=1/2`

Answer

B

Explanation

`(a^((n+1))+b^((n+1)))/(a^(n)+b^(n))` is GM between a and b
`rArr (a^((n+1))+b^((n+1)))/(a^(n)+b^(n))=a^(1/2)b^(1/2)`
`rArr [a^((n+1))+b^((n+1))]=(a^(n)+b^(n))(a^(1/2) b^(1/2))`
`rArr [a^((n+1))+b^((n+1))]=a^((n+1/2))+b^(1/2)+a^(1/2)b^((n+1/2))`
`rArr {a^((n+1))-a^((n+1/2))b^(1/2)}=a^(1/2)b^((n+1/2))-b^((n+1))`
`rArr a^((n+1/2))(a^(1/2)-b^(1/2))=b^((n+1/2))(a^(1/2)+b^(1/2))`
`rArr a^((n+1/2))=b^((n+1/2))" "[ :' (a^(1/2)-b^(1/2)) ne 0", since "a ne b]`
`rArr (a/b)^((n+1/2))=1=(a/b)^(0)`
`rArr n+1/2=0rArr n=-1/2`
Hence, `n=-1/2`.