{title2=$a_k+a_{n+1-k}=a_1+a_n$}

Reverse the finite list of <sequence terms> and add it to the original list. At position $k$,
$$
a_k+a_{n+1-k}
=\bigl(a_1+(k-1)d\bigr)+\bigl(a_1+(n-k)d\bigr)
=2a_1+(n-1)d=a_1+a_n.
$$
Every position therefore contributes the same value. Reversal preserves the <sum>, giving $2S_n=n(a_1+a_n)$ and the stated formula. The symmetry explains why the endpoint average represents the whole list; one does not need to add each entry separately.

\Image[/arithmetic-series-reversal-pairing.png]
{title=An <arithmetic progression> and its reverse make eight equal pairs, each of value thirty-five}
{description=Original positive-value example $7,10,13,16,19,22,25,28$. Blue bars show the original entries; yellow bars show the reversed entries. The combined <sum> is $280$, twice the original <sum> $140$. The algebraic argument also applies when some entries are negative.}

The reversal argument works for both <even> and <odd> lengths. For $2,5,8,11,14$, the two outer pairs each total sixteen and the middle entry is eight. Adding the list to its reverse counts that middle entry twice too, giving $2S_5=5\times16$ and $S_5=40$. No half-entry is required for an <odd> number of entries.


Pairing proof for an arithmetic series ( $a_{k} + a_{n + 1 - k} = a_{1} + a_{n}$ )

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