Arithmetic progression ($a_k=a_1+(k-1)d$)

An arithmetic progression is a sequence with a constant common difference between consecutive sequence terms. It has the form $a,a+r,\dots ,a+(m-1)r$ and is nonconstant when $r\ne 0$. Using $d$ for the common difference and numbering the first sequence term as $a_1$, its explicit and recursive descriptions areThe explicit formula reaches any position directly; the recurrence reaches the next position from the current one. The factor $k-1$ counts the steps after the first entry. For real $a_1,d$, a positive common difference makes the sequence increase, a negative one makes it decrease, and zero makes it constant.^[ref]

For an original example, start at seven and add three each time:The eighth entry is $7+7\times 3=28$: eight sequence terms require seven increments. This counting principle makes arithmetic progressions useful for regularly spaced positions, repeated equal additions, and iteration over evenly spaced integers. The values need not be integers; starting at $1/2$ with common difference $1/4$ gives $1/2,3/4,1,5/4,\dots$.

A geometric progression instead holds a ratio constant. In the displayed arithmetic progression, $10/7$ and $13/10$ differ even though $10-7=13-10=3$. Multiplying each entry by two would give a different kind of repeated update from adding three. Distinguishing the two rules matters before predicting later entries or adding the sequence terms.

Equal differences also appear in Nicomachus's treatment of arithmetic proportion. He describes successions of at least three numbers whose neighboring differences agree, then explains how regularly omitting counting numbers changes the common gap. This connects the progression to the arithmetic mean: a middle entry is equally far from its two neighbors.^[ref]

Colour each interval $[2^j,2^{j+1})\cap \mathbb{N}$ by the parity of $j$. Every infinite arithmetic progression of positive common difference d meets every sufficiently large such interval, since the interval has length greater than d. Both colours occur infinitely often. Thus neither colour class contains an infinite arithmetic progression, despite the finite-length conclusion of the Van der Waerden theorem.

If all residues of a modular arithmetic progression have integer representatives in an interval of diameter less than half the modulus, the representatives form an integer arithmetic progression. Consecutive differences lie in an interval shorter than the modulus, so the common modular difference has a unique possible integer representative.

The common difference of an arithmetic progression is the fixed difference between consecutive sequence terms. It gives a finite difference that is independent of position. Checking one adjacent pair finds its value only when the arithmetic progression property is already known; otherwise, other pairs can disagree. For example, the differences in $7,10,14$ are three and four, so the three entries do not form an arithmetic progression.

Starting at $a_1$ and repeatedly adding a nonzero common difference $d$, a proposed final value $z$ is reached at positionif and only if this expression is a positive integer. For $a_1=7$, $d=3$, and $z=28$, it gives $n=8$. The quotient $21/3=7$ counts increments, while the initial entry adds one to the number of sequence terms. The proposed value $27$ gives $n=1+20/3$, which is not an integer, so it is never reached by these updates. The test also handles decreasing arithmetic progressions; when $d=0$, the repeated value alone cannot determine a length.

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 $S_n=n{a}_1$.