Median (and consequently the mean) of an evenly-spaced list

Question

Why is it the case you can find the median (and consequently the mean) of an evenly-spaced list by taking the mean of opposite terms? Where opposite refers to opposite positions (e.g. first and last).

What intuition allows you to most easily reconcile this fact?

e.g. [1,2,...,49,50]

(1 + 50)/2 = 25.5 = mean = median

Answer 1

The mean is the median, since we can write the list as $m - k, m - (m - k + 1), \dots, m, m + 1, \dots, m +k$ if there are an odd number of terms, or similarly $m - (k - 0.5), m - (k - 1.5), \dots, m - 0.5, m + 0.5, \dots, m + (k - 0.5)$ if there are even number of terms, where $m$ is the median. In either case, the mean will be $m$, since the mean is \begin{align} \frac{(m - k) + (m - k + 1) + \dots + m + \dots + (m + k)}{2k + 1} & = \frac{(2k + 1)m + (-k) + k + (-k + 1) + (k - 1) + \dots}{2k + 1} \\ & = \frac{(2k + 1) m }{2k + 1} \\ & = m, \end{align} since we can group the $(2k + 1)$ $m$ terms together, and the rest of the terms in the numerator cancel out. Essentially the same thing happens if there are an even number of terms.

And the mean of the two outermost terms is also $m$, since that's just $((m - k) + (m + k)) / 2 = m$.

Answer 2

$\frac{1+2+\cdots 50}{50} = \frac{(1+50)\cdot 25}{50} = \frac{1+50}{2}$

Just note that the sum of the first and last element is equal to the sum of the second and second to last element (and so on)