Find the value of n so that ((a^(n+1) +b^(n+1) )/(a^n +b^n )) may become the A.M. between a and b.


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Question Number 3466 by Rasheed Soomro last updated on 13/Dec/15

$F i n d t h e v a l u e o f n s o t h a t$ $\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ $m a y b e c o m e t h e A . M . b e t w e e n$ $a a n d b .$
	Answered by prakash jain last updated on 13/Dec/15

	$n = 0 is solution .$ $\frac{a + b}{1 + 1} = \frac{a + b}{2}$
	Answered by RasheedSindhi last updated on 14/Dec/15

	$\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}} = \frac{a + b}{2}$ $2 (a^{n + 1} + b^{n + 1}) = (a + b) (a^{n} + b^{n}$ $= a^{n + 1} + {a b}^{n} + a^{n} b + b^{n + 1}$ $a^{n + 1} + b^{n + 1} = {a b}^{n} + a^{n} b$ $a^{n + 1} - a^{n} b + b^{n + 1} - {a b}^{n} = 0$ $a^{n} (a - b) - b^{n} (a - b) = 0$ $(a - b) (a^{n} - b^{n}) = 0$ $a - b = 0 ∣ a^{n} - b^{n} = 0$ $∙ a = b \Rightarrow n m a y b e a n y n u m b e r$ $i n t h i s c a s e .$ $∙ a^{n} = b^{n} \Rightarrow \frac{a^{n}}{b^{n}} = 1 [i f b \neq 0]$ ${(\frac{a}{b})}^{n} = {(\frac{a}{b})}^{0} \Rightarrow n = 0 [i f b \neq 0]$
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