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I have unified the 3 'M's

Due to the lack of response, I have decided to re-post my previous proof problem.

Consider a set of \(n\) consecutive terms from an arithmetic progression. Take any 2 terms and replace them by their average. Continue this until only one number remains. Prove that the mean of all end numbers (they need not be distinct) in the set is equal to the median and, for \(n \geq 4\), the mode.

Note by Sharky Kesa
2 years, 8 months ago

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If we translate all of the terms by any number, the mean, median, and mode are all translated by the same number. So WLOG we can assume that our arithmetic progression sums to 0 and consists of a bunch of numbers and their negatives. This actually is all I am using so far about the terms, not necessarily that they are in an arithmetic progression.

So it's clear that if we can get an end number \( x \) by one ordering of the terms, we can get \( -x \) by choosing the negatives of each term instead. So the end numbers are symmetric around \( 0 \), i.e. the set \( S \) of end numbers satisfies \( S = -S \). Any such set has mean and median equal to \( 0 \). I imagine this is what Calvin means by a symmetry argument.

Showing that the mode is \( 0 \) seems like it probably requires more work, and presumably requires us to actually use that the terms are in an arithmetic progression. Maybe someone can fill it in...? Or maybe I can, tomorrow.

Patrick Corn - 2 years, 6 months ago

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Yup, that's the symmetry argument for mean and median. Basically like you pointed out, instead of using \( a_i \) we use \( a_{n-i} \).

I'm not certain how to prove the mode part. I believe that there are \( \frac{ n! (n-1)! } { 2^{n-1} } \) ways to get the final value.

Calvin Lin Staff - 2 years, 6 months ago

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There is a very simple symmetry argument to this.

Hint: The final value is what you think it is (IE the value that makes the most sense).

Calvin Lin Staff - 2 years, 8 months ago

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I will try it!

Swapnil Das - 2 years, 8 months ago

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