Find-the-value-of-n-so-that-a-n-1-b-n-1-a-n-b-n-may-become-the-H-M-between-a-and-b-

Question Number 3615 by Rasheed Soomro last updated on 16/Dec/15

F i n d t h e v a l u e o f n s o t h a t

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

m a y b e c o m e t h e H . M . b e t w e e n

a a n d b .

Commented by prakash jain last updated on 16/Dec/15

n = - 1

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

Answered by Yozzii last updated on 16/Dec/15

H . M, h = \frac{1}{\frac{1}{2} (\frac{1}{a} + \frac{1}{b})} = \frac{2 a b}{a + b} . (a, b > 0)

I f h = \frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}

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

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

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

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

a^{n + 2} - {b a}^{n + 1} - {a b}^{n + 1} + b^{n + 2} = 0

a^{n + 1} (a - b) - b^{n + 1} (a - b) = 0

(a^{n + 1} - b^{n + 1}) (a - b) = 0

\Rightarrow a = b o r a^{n + 1} = b^{n + 1} (*) .

I f a \neq b, (*) i s t r u e i f f n = - 1 .

Commented by Rasheed Soomro last updated on 17/Dec/15

F o r b \neq 0

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

\Rightarrow n + 1 = 0 \Rightarrow n = - 1