Arithmetic progression

According to an anecdote of uncertain reliability,^[1] young Carl Friedrich Gauss, who was in primary school, reinvented the formula ${\displaystyle {\tfrac {n(n+1)}{2}}}$  for summing the integers from 1 through ${\displaystyle n}$ , for the case ${\displaystyle n=100}$ , by grouping the numbers from both ends of the sequence into pairs summing to 101 and multiplying by the number of pairs. 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.^[2] Similar rules were known in antiquity to Archimedes, Hypsicles and Diophantus;^[3] in China to Zhang Qiujian; in India to Aryabhata, Brahmagupta and Bhaskara II;^[4] and in medieval Europe to Alcuin,^[5] Dicuil,^[6] Fibonacci,^[7] Sacrobosco^[8] and to anonymous commentators of Talmud known as Tosafists.^[9]

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

${\displaystyle 2+5+8+11+14=40}$ 

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:

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

In the case above, this gives the equation:

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

This formula works for any real numbers ${\displaystyle a_{1}}$  and ${\displaystyle a_{n}}$ . For example: this

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

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

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

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

Rewriting the terms in reverse order:

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

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

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

This formula can be simplified as:

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

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

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

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

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

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

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

This is a generalization from the fact that the product of the progression ${\displaystyle 1\times 2\times \cdots \times n}$  is given by the factorial ${\displaystyle n!}$  and that the product

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

for positive integers ${\displaystyle m}$  and ${\displaystyle n}$  is given by

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

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

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

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

${\displaystyle \Gamma (z+2)=(z+1)\Gamma (z+1)=(z+1)z\Gamma ($z)}  ,

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

so that

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

for ${\displaystyle m}$  a positive integer and ${\displaystyle z}$  a positive complex number.

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

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

and, finally,

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

Example 1

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

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

Example 2

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

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

The standard deviation of any arithmetic progression can be calculated as

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

where ${\displaystyle n}$  is the number of terms in the progression and ${\displaystyle d}$  is the common difference between terms. The formula is very similar to the standard deviation of a discrete uniform distribution.

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.^[10] However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression.

• 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

• "Arithmetic series", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Weisstein, Eric W. "Arithmetic progression". MathWorld.
• Weisstein, Eric W. "Arithmetic series". MathWorld.