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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.

There are $N+1$ terms, namely $1,2,\dots ,N+1$. The arithmetic series formula givesPairing the first and last terms gives pair sum $N+2$, explaining both factors.

Pair the first and last terms of the arithmetic series. There are $500$ pairs, each summing to $1001$, so

There are $100$ terms in this arithmetic series, starting at $4$ and ending at $202$. Pairing the first and last terms, the second and penultimate terms, and so on, givesThe summand in the PDF is $2n$, not $2^n$.

The terms form an arithmetic progression with common difference $2$. There are $(55-5)/2+1=26$ terms. The arithmetic series formula gives