Is the mean always equal to the median of an arithmetic progression
e.g. for a set of consecutive integers $x, x+1, x+2, \dots,y-2, y-1, y$
The median is $(x+y)/2$ equals the mean $(x + x+1 +\dots+ y-1 + y)/(y-x+1)$
Say the endpoints are $x$ and $y$ in the arithmetic progression
In general is the median just the midpoint of these two numbers $- (x + y)/2$
the mean is the sum of the numbers divided by the number of terms
Best Answer
I will calculate the median and mean when there are $2n+1$ (odd number of) terms $$a, a+d, a+2d, ..., a+2nd $$ of the arithmetic progression, in the usual/standard notation. Then median is simply the middle term $$T_{n+1}=a+nd.$$ Also, sum of all terms $$S_{2n+1}=\dfrac{(2n+1)}{2}(a+(a+2nd))=(2n+1)(a+nd)$$ and therefore the mean (average) is given by $$\dfrac{S_{2n+1}}{2n+1}=a+nd.$$
I will leave you the same computations with even number of terms.