# 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. For instance, the sequence 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2.

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

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

and in general

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

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

## Sum

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 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:

${\displaystyle 2+5+8+11+14}$

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:

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

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:

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

### Derivation

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

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

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

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

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

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

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

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

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

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

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 2.18).

## 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

${\displaystyle a_{1}a_{2}\cdots a_{n}=d{\frac {a_{1}}{d}}d({\frac {a_{1}}{d}}+1)d({\frac {a_{1}}{d}}+2)\cdots d({\frac {a_{1}}{d}}+n-1)=d^{n}{\left({\frac {a_{1}}{d}}\right)}^{\overline {n}}=d^{n}{\frac {\Gamma \left(a_{1}/d+n\right)}{\Gamma \left(a_{1}/d\right)}},}$

where ${\displaystyle x^{\overline {n}}}$ denotes the rising factorial and ${\displaystyle \Gamma }$ denotes the Gamma function. (Note however that the formula is not valid when ${\displaystyle a_{1}/d}$ is a negative integer 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

${\displaystyle m\times (m+1)\times (m+2)\times \cdots \times (n-2)\times (n-1)\times n\,\!}$
${\displaystyle {\frac {n!}{(m-1)!}}.}$

Taking the example from above, the product of the terms of the arithmetic progression given by an = 3 + (n-1)(5) up to the 50th term is

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

## Standard deviation

The standard deviation of any arithmetic progression can be calculated via:

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

where ${\displaystyle n}$ is the number of terms in the progression, and ${\displaystyle d}$ is the common difference between terms

## Formulas at a Glance

Let

${\displaystyle a_{1}}$ is the first term of an arithmetic progression.
${\displaystyle a_{n}}$ is the nth term of an arithmetic progression.
${\displaystyle l}$ is the last term of an arithmetic progression.
${\displaystyle n}$ is the number of terms in the arithmetic progression.
${\displaystyle S_{n}}$ is the sum of n terms in the arithmetic progression.
${\displaystyle {\overline {n}}}$ is the mean value of arithmetic series.

then

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

## References

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