Solveeit Logo

Question

Question: If \(\frac { a ^ { n + 1 } + b ^ { n + 1 } } { a ^ { n } + b ^ { n } }\) be the A.M. of \(a\) and ...

If an+1+bn+1an+bn\frac { a ^ { n + 1 } + b ^ { n + 1 } } { a ^ { n } + b ^ { n } } be the A.M. of aa and bb , then n=n =

A

1

B

1- 1

C

0

D

None of these

Answer

0

Explanation

Solution

an+1+bn+1an+bn=a+b2\frac { a ^ { n + 1 } + b ^ { n + 1 } } { a ^ { n } + b ^ { n } } = \frac { a + b } { 2 }

\Rightarrow an+1abn+bn+1ban=0a ^ { n + 1 } - a b ^ { n } + b ^ { n + 1 } - b a ^ { n } = 0 \Rightarrow (ab)(anbn)=0( a - b ) \left( a ^ { n } - b ^ { n } \right) = 0

If anbn=0a ^ { n } - b ^ { n } = 0. Then (ab)n=1=(ab)0\left( \frac { a } { b } \right) ^ { n } = 1 = \left( \frac { a } { b } \right) ^ { 0 }.

Hencen=0n = 0.