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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

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  • $\begingroup$ Actually, we can have two medians, if not odd number of elements present $\endgroup$ – Yola Oct 25 '14 at 12:54
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First note that for calculate the median we need odd number of terms.
Let $$a, a+d, a+2d, ..., a+2nd $$ be your arithmetic progression of $2n+1$ (odd number) terms in usual notations. Then note that the median is $$T_{n+1}=a+nd.$$ Sum of all terms $$S_{2n+1}=\dfrac{(2n+1)}{2}(a+(a+2nd))=(2n+1)(a+nd).$$ Mean is given by $$\dfrac{S_{2n+1}}{2n+1}=a+nd.$$

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