Arithmetic series ($S_n={\sum }_{k=1}^n{a}_k$)

An arithmetic series adds the sequence terms of an arithmetic progression. Its $n$-th partial sum is $S_n={\sum }_{k=1}^n{a}_k$, with the empty sum convention $S_0=0$. For $n\ge 1$,Thus the total is the number of sequence terms multiplied by the arithmetic mean of the two endpoints. The formula also covers decreasing and constant arithmetic progressions.^[ref]

For the eight-entry example, the sum is $8(7+28)/2=140$, and the arithmetic mean of all eight entries is $17.5$. A decreasing example $11,9,7,5,3,1$ gives $6(11+1)/2=36$. Negative values cause no algebraic difficulty: $3,1,-1,-3,-5$ has sum $5(3-5)/2=-5$.

The triangular number $T_n=1+2+\cdots +n$ is the special case $a_1=d=1$, giving $n(n+1)/2$. More generally, an equal increase in each successive allocation makes the total grow as a quadratic function of the number of allocations. The eight-entry example hasIts first four entries add to $46$, its first eight to $140$, and its first twelve to $282$. The sequence terms lie on a straight line when plotted against their sequence indices, while the partial sums follow a quadratic function.

Reverse the finite list of sequence terms and add it to the original list. At position $k$,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.

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 $2S_5=5\times 16$ and $S_5=40$. No half-entry is required for an odd number of entries.

Adding one sequence term changes a partial sum by exactly that entry, so $S_{n+1}-S_n=a_{n+1}$. Taking another finite difference givesThus arithmetic sequence terms produce partial sums with constant second finite differences. Conversely, if these second finite differences are a fixed $d$ and $a_{n+1}=S_{n+1}-S_n$, then those entries form an arithmetic progression. For nonzero $d$, the displayed sum formula is a quadratic function; for $d=0$, it reduces to the linear expression $S_n=n{a}_1$.