I am not sure if this is solvable, I think I saw it somewhere if this question is not clear enough and missing information I would gladly remove it!
if the mean of an arithmetic consecutive numbers = $17$ and the median = $m$ $$mean = 17$$ $$median = m$$ what is $m-17$ = ?
Thanks for taking the time to read the question !
The hard way to do it that the sequence is $b, b+ d, b+ 2d, b+3d, ..... , b+ n*d$.
The mean is $\frac{\sum_{k=0}^n (b + dk)}{n+1} = \frac{(n+1)b + d\sum_{k=0}^n k}{n+1}$
$= b + \frac d{n+1}*\frac {n(n+1)}{2} = b + \frac {nd}2$.
The median is $b + $ halfway between $0$ and $n*d$ $= b+ \frac {nd}2$.
So mean - median = $0$.
The easy way to do it:
Let there be a sequence of $n = 2k + 1$ arithmetically sequential numbers. Let $m$ be the middle number. Let $m - k*d$ be the least and $m + kd$ be the greatest. $m$ is the middle and thus the median.
If you pair the terms up, the least ($m - k*d$) to the greatest ($m + k*d$) and the $j$th least $m - (k-j)d$ to the $j$th greatest ($m + (k-j)d$) then the average of each pair is $m$ and the average of all the terms is $m$.
So $m$ is both the median and the mean. So median - mean = $0$.
If the sequence has an even number $n = 2k$ terms, there won't be a middle term but the middle two terms will be $m - d/2$ and $m+d/2$ for some $m$ beings some mid point. The median with be $m $.
The terms all pair up with $m - d/2 - jd$ pairing with $m + d/2 + jd$. So the pairs add to $2m$ and average to $m$. So the mean is $m$>
So $m$ is both the median and the mean again.