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If $\displaystyle \frac{a^n + b^n}{a^{n-1} + b^{n-1}}$ , ( a != b and a,b,n are real numbers ) is the Arithmetic mean or Geometric mean between a and b then find the values of n for both means respectively.

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    Sorry the given answer is for A.M ,$n$= -1 and for G.M answer is 1/2.2010-09-27

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For the current version of the problem the answer is $n=1$ (for arithmetic mean) and $n=1/2$ (for geometric mean).

Consider first the case $ \frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a + b}{2} $ This equality implies $2a^n + 2b^n = a^{n} + ab^{n-1} + b^{n} + ba^{n-1}$. Hence $a^n + b^n = ab^{n-1} + ba^{n-1}$ and so $(a^{n-1} - b^{n-1})(a-b) = 0$. Hence (since $a \ne b$) we have $n=1$. It's easy to check that $n=1$ indeed satisfies the conditions.

Now consider the case $ \frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \sqrt{ab} $ This equality implies $a^n + b^n = a^{n-1/2}\sqrt{b} + \sqrt{a}b^{n-1/2}$. Hence $(a^{n-1/2} - b^{n-1/2})(\sqrt{a}-\sqrt{b}) = 0$. So (since $a \ne b$) we have $n=1/2$. It's easy to check that $n=1/2$ indeed satisfies the conditions.

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    With n = 0 we get the harmonic mean, and with n = ±∞ we get the maximum/minimum. These seems to be the only 5 points where OP's expression and the power mean $\sqrt[p]{\frac{a^p+b^p}2}$ coincide.2010-09-27
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Put $\rm\ A = a^{n-1},\ B = b^{n-1}\:$ below and the results follow immediately

$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ =\ \frac{a+b}2\ \iff\ (a-b)\ (A-B)\ =\ 0$

$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ = \ \ \sqrt{ab}\ \ \ \:\iff\ (\sqrt a - \sqrt b)\ (A\sqrt{a}-B\sqrt{b})\ =\ 0$