>Dndlp at 19:19, 10 June 2021

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{{short description|Sequence of numbers with constant differences between consecutive numbers}}
An '''Arithmetic progression (AP)''' or '''arithmetic sequence''' is a [[sequence]] of [[number]]s such that the difference between the consecutive terms is constant. 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 <math>a_1</math> and the common difference of successive members is ''d'', then the ''n''th term of the sequence (<math>a_n</math>) is given by:
:<math>\ a_n = a_1 + (n - 1)d</math>,

and in general

:<math>\ a_n = a_m + (n - m)d</math>.

A finite portion of an arithmetic progression is called a '''finite arithmetic progression''' and sometimes just called an arithmetic progression. The [[summation|sum]] of a finite arithmetic progression is called an '''arithmetic series'''.

==Sum==

<div class="thumb tright">
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{| style="background-color:white; width:220px;"
| 2 || + || 5 || + || 8 || + || 11 || + || 14 || = || 40
|-
| 14 || + || 11 || + || 8 || + || 5 || + || 2 || = || 40
|-
|colspan=11|<hr>
|-
| 16 || + || 16 || + || 16 || + || 16 || + || 16 || = || 80
|}
<div class="thumbcaption">
Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum.
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</div>
</div>
The [[Summation|sum]] of the members of a finite arithmetic progression is called an '''arithmetic series'''. For example, consider the sum:

:<math>2 + 5 + 8 + 11 + 14</math>

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:

:<math>\frac{n(a_1 + a_n)}{2}</math>

In the case above, this gives the equation:

:<math>2 + 5 + 8 + 11 + 14 = \frac{5(2 + 14)}{2} = \frac{5 \times 16}{2} = 40.</math>

This formula works for any real numbers <math>a_1</math> and <math>a_n</math>. For example:

:<math>\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}.</math>

=== Derivation ===
[[File:Animated proof for the formula giving the sum of the first integers 1+2+...+n.gif|thumb|Animated proof for the formula giving the sum of the first integers 1+2+...+n.]]
To derive the above formula, begin by expressing the arithmetic series in two different ways:
:<math> S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-2)d)+(a_1+(n-1)d)</math>

:<math> S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\cdots+(a_n-2d)+(a_n-d)+a_n.</math>

Adding both sides of the two equations, all terms involving ''d'' cancel:

:<math>\ 2S_n=n(a_1 + a_n).</math>

Dividing both sides by 2 produces a common form of the equation:

:<math> S_n=\frac{n}{2}( a_1 + a_n).</math>

An alternate form results from re-inserting the substitution: <math>a_n = a_1 + (n-1)d</math>:

:<math> S_n=\frac{n}{2}[ 2a_1 + (n-1)d].</math>

Furthermore, the mean value of the series can be calculated via: <math>S_n / n</math>:

:<math> \overline{a} =\frac{a_1 + a_n}{2}.</math>

The formula is very similar to the mean of a [[discrete uniform distribution]].

==Product==
The [[product (mathematics)|product]] of the members of a finite arithmetic progression with an initial element ''a''<sub>1</sub>, common differences ''d'', and ''n'' elements in total is determined in a closed expression

:<math>a_1a_2a_3\cdots a_n = a_1(a_1+d)(a_1+2d)...(a_1+(n-1)d)= \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)}</math>

where <math>\Gamma</math> denotes the [[Gamma function]]. The formula is not valid when <math>a_1/d</math> is negative or zero.

This is a generalization from the fact that the product of the progression <math>1 \times 2 \times \cdots \times n</math> is given by the [[factorial]] <math>n!</math> and that the product

:<math>m \times (m+1) \times (m+2) \times \cdots \times (n-2) \times (n-1) \times n </math>

for [[Natural number|positive integer]]s <math>m</math> and <math>n</math> is given by

:<math>\frac{n!}{(m-1)!}.</math>

===Derivation===

:<math>\begin{align}
a_1a_2a_3\cdots a_n &=\prod_{k=0}^{n-1} (a_1+kd) \\
&= \prod_{k=0}^{n-1} d\left(\frac{a_1}{d}+k\right) = 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 ) \\
&= 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{align}</math>
where <math>x^{\overline{n}}</math> denotes the [[Pochhammer symbol|rising factorial]].

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

:<math>\Gamma(z+2)=(z+1)\Gamma(z+1)=(z+1)z\Gamma(z)</math>,

:<math>\Gamma(z+3)=(z+2)\Gamma(z+2)=(z+2)(z+1)z\Gamma(z)</math>,

so that

:<math> \frac{\Gamma(z+m)}{\Gamma(z)} = \prod_{k=0}^{m-1}(z+k)</math>

for <math>m</math> a positive integer and <math>z</math> a positive complex number.

Thus, if <math>a_1/d > 0 </math>,

:<math>\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)}</math>,

and, finally,

:<math>a_1a_2a_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)} </math>

===Examples===

;Example 1
Taking the example <math> 3, 8, 13, 18, 23, 28, \ldots </math>, the product of the terms of the arithmetic progression given by <math>a_n = 3 + 5(n-1) </math> up to the 50<sup>th</sup> term is
:<math>P_{50} = 5^{50} \cdot \frac{\Gamma \left(3/5 + 50\right) }{\Gamma \left( 3 / 5 \right) } \approx 3.78438 \times 10^{98}. </math>

; Example 2
The product of the first 10 odd numbers <math>(1,3,5,7,9,11,13,15,17,19)</math> is given by
:<math> 1.3.5\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) } </math> = {{formatnum:654729075}}

==Standard deviation==

The standard deviation of any arithmetic progression can be calculated as

:<math> \sigma = |d|\sqrt{\frac{(n-1)(n+1)}{12}}</math>

where <math> n</math> is the number of terms in the progression and
<math> d</math> is the common difference between terms. The formula is very similar to the standard deviation of a [[discrete uniform distribution]].

==Intersections==
The [[Intersection (set theory)|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]].<ref>{{citation
| last = Duchet
| first = Pierre
| editor1-last = Graham | editor1-first = R. L.
| editor2-last = Grötschel | editor2-first = M. | editor2-link = Martin Grötschel
| editor3-last = Lovász | editor3-first = L.
| contribution = Hypergraphs
| location = Amsterdam
| mr = 1373663
| pages = 381–432
| publisher = Elsevier
| title = Handbook of combinatorics, Vol. 1, 2
| year = 1995
}}. See in particular Section 2.5, "Helly Property", [https://books.google.com/books?id=5Y9NCwlx63IC&pg=PA393 pp. 393–394].</ref> However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression.

==History==
According to an anecdote of uncertain reliability,<ref name="hayesreckoning">{{cite journal|author=Hayes|first=Brian|date=2006|title=Gauss's Day of Reckoning|url=https://www.americanscientist.org/article/gausss-day-of-reckoning |url-status=live|journal=[[American Scientist]]|volume=94|issue=3|page=200|doi=10.1511/2006.59.200|archive-url=https://web.archive.org/web/20120112140951/http://www.americanscientist.org/issues/id.3483,y.0,no.,content.true,page.1,css.print/issue.aspx|archive-date=12 January 2012|access-date=16 October 2020}}</ref> young [[Carl Friedrich Gauss#Anecdotes|Carl Friedrich Gauss]] in primary school reinvented this method to compute the sum of the integers from 1 through 100, by multiplying {{math|{{sfrac|''n''|2}}}} pairs of numbers in the sum by the values of each pair {{math|''n'' + 1}}. However, regardless of the truth of this story, Gauss was not the first to discover this formula, and some find it likely that its origin goes back to the [[Pythagoreans]] in the 5th century BC.<ref>Høyrup, J. The “Unknown Heritage”: trace of a forgotten locus of mathematical sophistication. Arch. Hist. Exact Sci. 62, 613–654 (2008). https://doi.org/10.1007/s00407-008-0025-y</ref> Similar rules were known in antiquity to [[Archimedes]], [[Hypsicles]] and [[Diophantus]];<ref>{{cite book
| title = Analysis, analytische Geometrie
| url = https://books.google.com/books?id=9dJ_F4lCXTQC&hl
| url-access = limited
| author = Tropfke, Johannes
| publisher = Walter de Gruyter
| year = 1924
| isbn = 978-3-11-108062-8
| pages = 3-15}}</ref> in China to [[Zhang Qiujian]]; in India to [[Aryabhata]], [[Brahmagupta]] and [[Bhaskara II]];<ref>{{cite book
| title = Arithmetik und Algebra
| url = https://books.google.com/books?id=7UW0DwAAQBAJ&hl
| url-access = limited
| author = Tropfke, Johannes
| publisher = Walter de Gruyter
| year = 1979
| isbn = 978-3-11-004893-3
| pages = 344-354}}</ref> and in medieval Europe to [[Alcuin]],<ref name=a>[https://www.jstor.org/stable/3620384 Problems to Sharpen the Young], John Hadley and David Singmaster, ''The Mathematical Gazette'', '''76''', #475 (March 1992), pp. 102–126.</ref> [[Dicuil]], <ref>Ross, H.E. & Knott,B.I (2019) Dicuil (9th century) on triangular and square numbers, ''British Journal for the History of Mathematics'', 34:2, 79-94, https://doi.org/10.1080/26375451.2019.1598687</ref> [[Fibonacci]], <ref>
{{cite book
| title = Fibonacci's Liber Abaci
| url = https://archive.org/details/fibonaccislibera00sigl
| url-access = limited
| author = Sigler, Laurence E. (trans.)
| publisher = Springer-Verlag
| year = 2002
| isbn = 0-387-95419-8
| pages = [https://archive.org/details/fibonaccislibera00sigl/page/n260 259]–260}}</ref> [[Sacrobosco]]<ref>
{{cite book
| title = Sourcebook in the Mathematics of Medieval Europe and North Africa
| url = https://books.google.com/books?id=39waDQAAQBAJ&hl
| url-access = limited
| author = Katz, Victor J. (edit.)
| publisher = Princeton University Press
| year = 2016
| isbn = 9780691156859
| pages = 91,257}}</ref>
and to anonymous commentators of Talmud known as [[Tosafot|Tosafists]].<ref>Stern, M. (1990). 74.23 A Mediaeval Derivation of the Sum of an Arithmetic Progression. The Mathematical Gazette, 74(468), 157-159. doi:10.2307/3619368</ref>

==See also==
* [[Geometric progression]]
* [[Harmonic progression (mathematics)|Harmonic progression]]
* [[Triangular number]]
* [[Arithmetico-geometric sequence]]
* [[Inequality of arithmetic and geometric means]]
* [[Primes in arithmetic progression]]
* [[Linear difference equation]]
* [[Generalized arithmetic progression]], a set of integers constructed as an arithmetic progression is, but allowing several possible differences
* [[Integer triangle#Heronian triangles with sides in arithmetic progression|Heronian triangles with sides in arithmetic progression]]
* [[Problems involving arithmetic progressions]]
* [[Utonality]]

== References ==
<references/>

==External links==
* {{springer|title=Arithmetic series|id=p/a013370}}
* {{MathWorld|urlname=ArithmeticProgression|title=Arithmetic progression}}
* {{MathWorld|urlname=ArithmeticSeries|title=Arithmetic series}}

{{Series (mathematics)}}

{{DEFAULTSORT:Arithmetic Progression}}
[[Category:Arithmetic series]]
[[Category:Articles containing proofs]]
[[Category:Sequences and series]]

Arithmetic progression - Revision history

>Dndlp at 19:19, 10 June 2021