# Arithmetic progression

(Redirected from Arithmetic series)

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 $a_1$ 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, the members (terms) will grow towards positive infinity.
• Negative, the members (terms) will grow towards negative infinity.

## Sum

This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.
 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: $S_n / n$:

$\overline{n} =\frac{a_1 + a_n}{2}.$

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

$a_1a_2\cdots a_n = d \frac{a_1}{d} d (\frac{a_1}{d}+1)d (\frac{a_1}{d}+2)\cdots d (\frac{a_1}{d}+n-1)=d^n {\left(\frac{a_1}{d}\right)}^{\overline{n}} = d^n \frac{\Gamma \left(a_1/d + n\right) }{\Gamma \left( a_1 / d \right) },$

where $x^{\overline{n}}$ denotes the rising factorial and $\Gamma$ denotes the Gamma function. (Note however that the formula is not valid when $a_1/d$ is a negative integer 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

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

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

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

## Standard deviation

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

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

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

## Formulas at a Glance

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.
$\overline{n}$ 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. $S_n=\frac{n(a_1 + a_n)}{2}$
5. $\overline{n}$ = $S_n / n$
6. $\overline{n} =\frac{a_1 + a_n}{2}.$