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My question is ,

lets say we have the numbers 1,2,3,.....,n numbers . 2 consecutive numbers are removed. The avg of the numbers after the removal of numbers be m

let the numbers removed be k,k+1

can we come up with a formula that relates m,n,k

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up vote 2 down vote accepted

The average of all the integers from 1 to $n$ is given by $$ \frac{\frac{1}{2}n(n+1)}{n}, $$ or simply $$ \frac{1}{2}(n+1). $$

The sum of the integers from 1 to $n$ with $k$ and $k+1$ removed is given by $$ \frac{n(n+1)}{2} - k - (k+1), $$ which we can clean up slightly to $$ \frac{1}{2}(n^2 + n - 4k - 2) $$ The expression above is the sum of $n-2$ numbers, so the average of these numbers is $$ \frac{\frac{1}{2}(n^2 + n - 4k - 2)}{n-2}, $$ or simply $$ \frac{n^2 + n - 4k - 2}{2(n-2)}. $$ Thus, the amount by which the average decreased is given by $$ \frac{1}{2}(n+1) - \frac{n^2 + n - 4k - 2}{2(n-2)}, $$ which simplifies to $$ \frac{2k - n}{n-2}. $$

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The last formula is the amount by which the average decreased. If one removes two values on the lower half of the spectrum, then the average should increase (i.e. decrease by a negative amount). – Austin Mohr Nov 16 '11 at 7:32
Oh, of course :-) I should get more sleep these days. – Listing Nov 16 '11 at 7:34

In general, if you have $n$ numbers whose mean is $\mu$, the sum of those numbers is $n \mu$. If you then remove $m$ of those numbers whose mean is $\nu$ (and thus whose sum is $m \mu$), the remaining $n-m$ numbers have mean $\displaystyle\frac{n\mu - m \nu}{n-m} = \mu + \frac{m (\mu - \nu)}{n-m}$.

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This answer, when contrasted with mine, is like a case study in how generality can lead to elegance. – Austin Mohr Nov 16 '11 at 18:34

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