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Arithmetic progression

Sequence of equally spaced numbers

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Source updated September 8, 2026

An arithmetic progression, arithmetic sequence or linear sequence is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence. The constant difference is called common difference of that arithmetic progression. For instance, the sequence 5, 7, 9, 11, 13, 15, ... is an arithmetic progression with a common difference of 2.

If the initial term of an arithmetic progression is $a_{1}$ and the common difference of successive members is $d$ , then the $n$ -th term of the sequence ( $a_{n}$ ) is given by

$a_{n}=a_{1}+(n-1)d.$

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.

01History

According to an anecdote of uncertain reliability, in primary school Carl Friedrich Gauss reinvented the formula ${\tfrac {n(n+1)}{2}}$ for summing the integers from 1 through $n$ , for the case $n=100$ , by grouping the numbers from both ends of the sequence into pairs summing to 101 and multiplying by the number of pairs. Regardless of the truth of this story, Gauss was not the first to discover this formula. Similar rules were known in antiquity to Archimedes, Hypsicles and Diophantus; in China to Zhang Qiujian; in India to Aryabhata, Brahmagupta and Bhaskara II; and in medieval Europe to Alcuin, Dicuil, Fibonacci, Sacrobosco, and anonymous commentators of Talmud known as Tosafists. Some find it likely that its origin goes back to the Pythagoreans in the 5th century BC.

02Sum

The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum:

$2+5+8+11+14=40$

This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2:

${\frac {n(a_{1}+a_{n})}{2}}$

In the case above, this gives the equation:

$2+5+8+11+14={\frac {5(2+14)}{2}}={\frac {5\times 16}{2}}=40.$

This formula works for any arithmetic progression of real numbers beginning with $a_{1}$ and ending with $a_{n}$ . For example,

$\left(-{\frac {3}{2}}\right)+\left(-{\frac {1}{2}}\right)+{\frac {1}{2}}={\frac {3\left(-{\frac {3}{2}}+{\frac {1}{2}}\right)}{2}}=-{\frac {3}{2}}.$

Derivation

To derive the above formula, begin by expressing the arithmetic series in two different ways:

$S_{n}=a+a_{2}+a_{3}+\dots +a_{(n-1)}+a_{n}$

$S_{n}=a+(a+d)+(a+2d)+\dots +(a+(n-2)d)+(a+(n-1)d).$

Rewriting the terms in reverse order:

$S_{n}=(a+(n-1)d)+(a+(n-2)d)+\dots +(a+2d)+(a+d)+a.$

Adding the corresponding terms of both sides of the two equations and halving both sides:

$S_{n}={\frac {n}{2}}[2a+(n-1)d].$

This formula can be simplified as:

${\begin{aligned}S_{n}&={\frac {n}{2}}[a+a+(n-1)d].\\&={\frac {n}{2}}(a+a_{n}).\\&={\frac {n}{2}}({\text{initial term}}+{\text{last term}}).\end{aligned}}$

Furthermore, the mean value of the series can be calculated via: $S_{n}/n$ :

${\overline {a}}={\frac {a_{1}+a_{n}}{2}}.$

The formula is essentially the same as the formula for the mean of a discrete uniform distribution, interpreting the arithmetic progression as a set of equally probable outcomes.

03Product

The product of the members of a finite arithmetic progression with an initial element a₁, common differences d, and n elements in total is determined in a closed expression

${\begin{aligned}a_{1}a_{2}a_{3}\cdots a_{n}&=a_{1}(a_{1}+d)(a_{1}+2d)\cdots (a_{1}+(n-1)d)\\[1ex]&=\prod _{k=0}^{n-1}(a_{1}+kd)=d^{n}{\frac {\Gamma {\left({\frac {a_{1}}{d}}+n\right)}}{\Gamma {\left({\frac {a_{1}}{d}}\right)}}}\end{aligned}}$

where $\Gamma$ denotes the Gamma function. The formula is not valid when $a_{1}/d$ is negative or zero.

This is a generalization of the facts that the product of the progression $1\times 2\times \cdots \times n$ is given by the factorial $n!$ and that the product

$m\times (m+1)\times (m+2)\times \cdots \times (n-2)\times (n-1)\times n$

for positive integers $m$ and $n$ is given by

${\frac {n!}{(m-1)!}}.$

Derivation

${\begin{aligned}a_{1}a_{2}a_{3}\cdots a_{n}&=\prod _{k=0}^{n-1}(a_{1}+kd)\\[2pt]&=\prod _{k=0}^{n-1}d\left({\frac {a_{1}}{d}}+k\right)\\[2pt]&=d\left({\frac {a_{1}}{d}}\right)d\left({\frac {a_{1}}{d}}+1\right)d\left({\frac {a_{1}}{d}}+2\right)\cdots d\left({\frac {a_{1}}{d}}+(n-1)\right)\\[2pt]&=d^{n}\prod _{k=0}^{n-1}\left({\frac {a_{1}}{d}}+k\right)=d^{n}{\left({\frac {a_{1}}{d}}\right)}^{\overline {n}}\end{aligned}}$

where $x^{\overline {n}}$ denotes the rising factorial.

By the recurrence formula $\Gamma (z+1)=z\Gamma (z)$ , valid for a complex number $z>0$ ,

$\Gamma (z+2)=(z+1)\Gamma (z+1)=(z+1)z\Gamma (z)$ ,

$\Gamma (z+3)=(z+2)\Gamma (z+2)=(z+2)(z+1)z\Gamma (z)$ ,

so that

${\frac {\Gamma (z+m)}{\Gamma (z)}}=\prod _{k=0}^{m-1}(z+k)$

for $m$ a positive integer and $z$ a positive complex number.

Thus, if $a_{1}/d>0$ ,

$\prod _{k=0}^{n-1}\left({\frac {a_{1}}{d}}+k\right)={\frac {\Gamma {\left({\frac {a_{1}}{d}}+n\right)}}{\Gamma {\left({\frac {a_{1}}{d}}\right)}}},$

and, finally,

$a_{1}a_{2}a_{3}\cdots a_{n}=d^{n}\prod _{k=0}^{n-1}\left({\frac {a_{1}}{d}}+k\right)=d^{n}{\frac {\Gamma {\left({\frac {a_{1}}{d}}+n\right)}}{\Gamma {\left({\frac {a_{1}}{d}}\right)}}}$

Examples

Example 1

Taking the example $3,8,13,18,23,28,\ldots$ , the product of the terms of the arithmetic progression given by $a_{n}=3+5(n-1)$ up to the 50th term is

$P_{50}=5^{50}\cdot {\frac {\Gamma \left(3/5+50\right)}{\Gamma \left(3/5\right)}}\approx 3.78438\times 10^{98}.$

Example 2

The product of the first 10 odd numbers $(1,3,5,7,9,11,13,15,17,19)$ is given by

$1\cdot 3\cdot 5\cdots 19=\prod _{k=0}^{9}(1+2k)=2^{10}\cdot {\frac {\Gamma \left({\frac {1}{2}}+10\right)}{\Gamma \left({\frac {1}{2}}\right)}}$ = 654,729,075

04Standard deviation

The standard deviation of any arithmetic progression is

$\sigma =|d|{\sqrt {\frac {(n-1)(n+1)}{12}}}$

where $n$ is the number of terms in the progression and $d$ is the common difference between terms. The formula is essentially the same as the formula for the standard deviation of a discrete uniform distribution, interpreting the arithmetic progression as a set of equally probable outcomes.

05Intersections

The intersection of any two doubly infinite arithmetic progressions is either empty or another arithmetic progression, which can be found using the Chinese remainder theorem. If each pair of progressions in a family of doubly infinite arithmetic progressions have a non-empty intersection, then there exists a number common to all of them; that is, infinite arithmetic progressions form a Helly family. However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression.

06Amount of arithmetic subsets of length k of the set {1,...,n}

Let $a(n,k)$ denote the number of arithmetic subsets of length $k$ one can make from the set $\{1,\cdots ,n\}$ and let $\phi (\eta ,\kappa )$ be defined as:

$\phi (\eta ,\kappa )={\begin{cases}0&{\text{if }}\kappa \mid \eta \\\left(\left[\eta \;({\text{mod }}\kappa )\right]-2\right)\left(\kappa -\left[\eta \;({\text{mod }}\kappa )\right]\right)&{\text{if }}\kappa \not \mid \eta \\\end{cases}}$

Then:

${\begin{aligned}a(n,k)&={\frac {1}{2(k-1)}}\left(n^{2}-(k-1)n+(k-2)+\phi (n+1,k-1)\right)\\&={\frac {1}{2(k-1)}}\left((n-1)(n-(k-2))+\phi (n+1,k-1)\right)\end{aligned}}$

As an example, if ${\textstyle (n,k)=(7,3)$ , one expects ${\textstyle a(7,3)=9$ arithmetic subsets and, counting directly, one sees that there are 9; these are ${\textstyle \{1,2,3\},\{2,3,4\},\{3,4,5\},\{4,5,6\},\{5,6,7\},\{1,3,5\},\{3,5,7\},\{2,4,6\},\{1,4,7\}.$

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Sources and credits

This article is adapted from the Wikipedia article “Arithmetic progression”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

Images, from Wikimedia Commons:

Arithmetic progression.svg by Cmglee, CC BY-SA 4.0
Animated proof for the formula giving the sum of the first integers 1+2+...+n.gif by Vincent Pantaloni, CC BY-SA 4.0

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Arithmetic progression

01History

02Sum

Derivation

03Product

Derivation

Examples

04Standard deviation

05Intersections

06Amount of arithmetic subsets of length k of the set {1,...,n}

Sources and credits

Related topics

Pythagorean triple

Geometric progression

Harmonic progression (mathematics)