Arithmetic progression

According to an anecdote of uncertain reliability,{{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}} in primary school Carl Friedrich Gauss reinvented the formula $\backslash tfrac\{n(n+1)\}\{2\}$ 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;{{cite book |author=Tropfke, Johannes |url=https://books.google.com/books?id=9dJ_F4lCXTQC |title=Analysis, analytische Geometrie |publisher=Walter de Gruyter |year=1924 |isbn=978-3-11-108062-8 |pages=3–15 |url-access=limited}} in China to Zhang Qiujian; in India to Aryabhata, Brahmagupta and Bhaskara II;{{cite book |author=Tropfke, Johannes |url=https://books.google.com/books?id=7UW0DwAAQBAJ |title=Arithmetik und Algebra |publisher=Walter de Gruyter |year=1979 |isbn=978-3-11-004893-3 |pages=344–354 |url-access=limited}} and in medieval Europe to Alcuin,[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. Dicuil,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 Fibonacci,

{{cite book |author=Sigler, Laurence E. (trans.) |url=https://archive.org/details/fibonaccislibera00sigl |title=Fibonacci's Liber Abaci |publisher=Springer-Verlag |year=2002 |isbn=0-387-95419-8 |pages=[https://archive.org/details/fibonaccislibera00sigl/page/n260 259]–260 |url-access=limited}} Sacrobosco,

{{cite book |author=Katz, Victor J. (edit.) |url=https://books.google.com/books?id=39waDQAAQBAJ |title=Sourcebook in the Mathematics of Medieval Europe and North Africa |publisher=Princeton University Press |year=2016 |isbn=9780691156859 |pages=91,257 |url-access=limited}} and anonymous commentators of Talmud known as Tosafists.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 Some find it likely that its origin goes back to the Pythagoreans in the 5th century BC.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

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:

:$\backslash frac\{n(a\_1\; +\; a\_n)\}\{2\}$

In the case above, this gives the equation:

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

This formula works for any arithmetic progression of real numbers beginning with $a\_1$ and ending with $a\_n$. For example,

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

File:Animated proof for the formula giving the sum of the first integers 1+2+...+n.gif

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

:$S\_n=a+a\_2+a\_3+\backslash dots+a\_\{(n-1)\}\; +a\_n$

:$S\_n=a+(a+d)+(a+2d)+\backslash dots+(a+(n-2)d)+(a+(n-1)d).$

Rewriting the terms in reverse order:

:$S\_n=(a+(n-1)d)+(a+(n-2)d)+\backslash dots+(a+2d)+(a+d)+a.$

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

:$S\_n=\backslash frac\{n\}\{2\}[2a\; +\; (n-1)d].$

This formula can be simplified as:

:$\backslash begin\{align\}$

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{align}

Furthermore, the mean value of the series can be calculated via: $S\_n\; /\; n$:

:$\backslash overline\{a\}\; =\backslash frac\{a\_1\; +\; a\_n\}\{2\}.$

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.

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

:$\backslash begin\{align\}$

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{align}

where $\backslash Gamma$ 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 $1\; \backslash times\; 2\; \backslash times\; \backslash cdots\; \backslash times\; n$ is given by the factorial $n!$ and that the product

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

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

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

:$\backslash begin\{align\}$

a_1a_2a_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{align}

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

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

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

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

so that

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

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

Thus, if $a\_1/d\; >\; 0$,

:$\backslash prod\_\{k=0\}^\{n-1\}\; \backslash left(\backslash frac\{a\_1\}\{d\}+k\backslash right)=\; \backslash frac\{\backslash Gamma\{\backslash left(\backslash frac\{a\_1\}\{d\}\; +\; n\backslash right)\}\}\{\backslash Gamma\{\backslash left(\; \backslash frac\{a\_1\}\{d\}\; \backslash right)\}\},$

and, finally,

:$a\_1a\_2a\_3\backslash cdots\; a\_n\; =\; d^n\backslash prod\_\{k=0\}^\{n-1\}\; \backslash left(\backslash frac\{a\_1\}\{d\}+k\backslash right)\; =\; d^n\; \backslash frac\{\backslash Gamma\{\backslash left(\backslash frac\{a\_1\}\{d\}\; +\; n\backslash right)\}\}\{\backslash Gamma\{\backslash left(\; \backslash frac\{a\_1\}\{d\}\; \backslash right)\}\}$

;Example 1

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

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

; Example 2

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

:$1\backslash cdot\; 3\backslash cdot\; 5\backslash cdots\; 19\; =\backslash prod\_\{k=0\}^\{9\}\; (1+2k)\; =\; 2^\{10\}\; \backslash cdot\; \backslash frac\{\backslash Gamma\; \backslash left(\backslash frac\{1\}\{2\}\; +\; 10\backslash right)\; \}\{\backslash Gamma\; \backslash left(\; \backslash frac\{1\}\{2\}\; \backslash right)\; \}$ = {{formatnum:654729075}}

The standard deviation of any arithmetic progression is

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

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.

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.{{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]. However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression.

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

$\backslash phi(\backslash eta,\; \backslash kappa)\; =\; \backslash begin\{cases\}$

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:

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

• Geometric progression
• 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
• Heronian triangles with sides in arithmetic progression
• Problems involving arithmetic progressions
• Utonality
• Polynomials calculating sums of powers of arithmetic progressions

{{Reflist}}

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

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Category:Arithmetic series

Category:Articles containing proofs

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