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.
A simple algebra problem
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1What do you mean by "Arithmetic mean / Geometric mean"? There is an $n$ for each kind of mean? – 2010-09-27
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1This 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. – 2010-09-27
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0@Debanjan: What Kenny TM, was thinking was perhaps, whether quantity given is a A.M or a G.M or both. – 2010-09-27
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1@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. – 2010-09-27
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0@Debanjan: Hey what did "/" stand for in your question. I thought it stood for divides. If not then please tell me i shall edit again. – 2010-09-27
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0@ KennyTM: My bad.It's now working correct :) Btw I am inquisitive to know what application are you using to format the equation ? – 2010-09-27
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0If $a=b$ then any $n$ satisfies your conditions. – 2010-09-27
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0@Debanjan: Is $n \in \mathbb{N}$. You are saying the answer to be -1/2, -1. So you should mentioned that $n$ need not be an natural number. – 2010-09-27
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0I edited, I think it's stands good now,anyways this is an objective type problem and the problem-setter haven't mentioned these restrictions. – 2010-09-27
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1@Deb: (1) http://meta.math.stackexchange.com/questions/107/what-should-go-in-the-math-stackexchange-faq/117#117; (2) If so then your answers (-1/2, -1) are incorrect. We get the AM with n = 1, not n = -1/2. – 2010-09-27
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0Sorry the given answer is for A.M , n = -1 and for G.M answer is 1/2. – 2010-09-27
2 Answers
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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0This is very correct,since your answers matches the options given there, I believes there is some printing mistakes in the printed solutions. – 2010-09-27
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1With 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
Put $\rm\ A = a^{n-1},\ B = b^{n-1}\:$ below and the results follow immediately
$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ =\ \frac{a+b}2\ \iff\ (a-b)\ (A-B)\ =\ 0$
$\rm\displaystyle\quad \frac{a\:A+b\:B}{A+B}\ = \ \ \sqrt{ab}\ \ \ \:\iff\ (\sqrt a - \sqrt b)\ (A\sqrt{a}-B\sqrt{b})\ =\ 0$