# A simple algebra problem

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.

-
What do you mean by "Arithmetic mean / Geometric mean"? There is an $n$ for each kind of mean? – kennytm Sep 27 '10 at 9:52
This is not my homework,this comes from my test paper, the solution given is is -1/2 ans -1 receptively. Arithmetic mean and geometric mean are seems similar to the concept of A.M and G.M respectively of the sequence and series chapter in higher algebra. – Quixotic Sep 27 '10 at 9:58
@Debanjan: What Kenny TM, was thinking was perhaps, whether quantity given is a A.M or a G.M or both. – anonymous Sep 27 '10 at 9:59
@Debanjan: For $n=1$ clearly the quantity represents an A.M between the numbers $a$ and $b$. And it is still unclear as to what you want. – anonymous Sep 27 '10 at 10:01
@Deb: (1) meta.math.stackexchange.com/questions/107/…; (2) If so then your answers (-1/2, -1) are incorrect. We get the AM with n = 1, not n = -1/2. – kennytm Sep 27 '10 at 10:13

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.

-
This is very correct,since your answers matches the options given there, I believes there is some printing mistakes in the printed solutions. – Quixotic Sep 27 '10 at 10:21
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. – kennytm Sep 27 '10 at 18:11

Put $\rm\ A = a^{n-1},\ B = b^{n-1}\:$ below and the results follow immediately

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

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

-