# Arithmetic progression

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second minus the first. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with common difference of 2. Visual proof of the derivation of arithmetic progression formulas – the faded blocks are a rotated copy of the arithmetic progression

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

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

and in general

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

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.

The behavior of the arithmetic progression depends on the common difference d. If the common difference is:

• positive, then the members (terms) will grow towards positive infinity;
• negative, then the members (terms) will grow towards negative infinity.

## Sum

 2 + 5 + 8 + 11 + 14 = 40 14 + 11 + 8 + 5 + 2 = 40 16 + 16 + 16 + 16 + 16 = 80

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.

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

$2+5+8+11+14$

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 real numbers $a_{1}$  and $a_{n}$ . 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:

$S_{n}=a_{1}+(a_{1}+d)+(a_{1}+2d)+\cdots +(a_{1}+(n-2)d)+(a_{1}+(n-1)d)$
$S_{n}=(a_{n}-(n-1)d)+(a_{n}-(n-2)d)+\cdots +(a_{n}-2d)+(a_{n}-d)+a_{n}.$

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

$\ 2S_{n}=n(a_{1}+a_{n}).$

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

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

An alternate form results from re-inserting the substitution: $a_{n}=a_{1}+(n-1)d$ :

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

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 very similar to the mean of a discrete uniform distribution.

In 499 AD Aryabhata, a prominent mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, gave this method in the Aryabhatiya (section 2.19).

According to an anecdote, young Carl Friedrich Gauss reinvented this method to compute the sum 1+2+3+...+99+100 for a punishment in primary school.

## Product

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

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

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

This is a generalization from the fact 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)\\&=\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 $x^{\overline {n}}$  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 $3,8,13,18,23,28,\cdots$ , 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

$1.3.5\cdots 19=\prod _{k=0}^{9}(1+2k)=2^{10}\cdot {\frac {\Gamma \left({\frac {1}{2}}+10\right)}{\Gamma \left({\frac {1}{2}}\right)}}=654729075$

## Standard deviation

The standard deviation of any arithmetic progression can be calculated as

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

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

## Intersections

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.

## In Programming

### Python

Arange and linspace from numpy are used to generate AP sequence.

arange(start/intitial term, stop/final term, step/common difference)

linspace(start/initial term, stop/final term, no. of terms)

## Summary of formulae

If

$a_{1}$  is the first term of an arithmetic progression.
$a_{n}$  is the nth term of an arithmetic progression.
$d$  is the difference between terms of the arithmetic progression.
$n$  is the number of terms in the arithmetic progression.
$S_{n}$  is the sum of n terms in the arithmetic progression.
${\overline {a}}$  is the mean value of arithmetic series.

then

1. $\ a_{n}=a_{1}+(n-1)d,$
2. $\ a_{n}=a_{m}+(n-m)d.$
3. $S_{n}={\frac {n}{2}}[2a_{1}+(n-1)d].$
4. $S_{n}={\frac {n}{2}}(a_{1}+a_{n}).$
5. ${\overline {a}}={\frac {a_{1}+a_{n}}{2}}.$
6. $d={\frac {a_{m}-a_{n}}{m-n}};m\neq n$ .