# Effect of removal of numbers while calculating average

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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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 $$\frac{2k - n}{n-2}.$$
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 $\displaystyle\frac{n\mu - m \nu}{n-m} = \mu + \frac{m (\mu - \nu)}{n-m}$.