Given-that-a-n-1-b-n-1-a-n-b-n-is-AM-between-a-and-b-where-a-b-a-b-0-find-out-the-value-of-n-

Question Number 28327 by Rasheed.Sindhi last updated on 24/Jan/18

Given that \frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}} is AM between a

and b, where a \neq b \land a, b \neq 0; find out the value of n .

Answered by mrW2 last updated on 24/Jan/18

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

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

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

a^{n + 1} + b^{n + 1} = {a b}^{n} + {b a}^{n}

a^{n} (a - b) = b^{n} (a - b)

s i n c e a \neq b, a - b \neq 0 a n d \frac{a}{b} \neq 1

\Rightarrow a^{n} = b^{n}

\Rightarrow {(\frac{a}{b})}^{n} = 1

\Rightarrow n = \frac{\log 1}{\log (\frac{a}{b})} = \frac{0}{\log (\frac{a}{b})} = 0

Commented by Rasheed.Sindhi last updated on 25/Jan/18

T h α n k s a lot Sir!