# Arithmetic progression

In mathematics, an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 3, 5, 7, 9, 11, 13... is an arithmetic progression with common difference 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 is given by:

$\displaystyle \ a_n = a_1 + (n - 1)d,$

and in general

$\displaystyle \ a_n = a_m + (n - m)d.$

## Sum (arithmetic series)

The sum of the components of an arithmetic progression is called an arithmetic series.

### Calculating the value of an arithmetic series

The value of an arithmetic series consisting of n terms $\displaystyle a_1,a_2,\dots,a_n$ is given by

$\displaystyle S_n = a_1+a_2+\dots+a_n=\frac{n( a_1 + a_n)}{2} =\frac{n[ 2a_1 + (n-1)d ]}{2}.$

Intuitively, this formula can be derived by realizing that the sum of the first and last terms in the series is the same as the sum of the second and second to last terms, and so forth, and that there are roughly $\displaystyle n/2$ such sums in the series. A version of this formula appears in the Liber Abaci (1202, ch. II.12) of Leonardo of Pisa (commonly known as Fibonacci). An often-told story is that Carl Friedrich Gauss rediscovered this formula when his third grade teacher asked the class to find the sum of the first 100 numbers, and he instantly computed the answer (5050).

A different way to get the result, that avoids the fuzziness of the previous method when the number of terms is odd, is to think in terms of averages. The value of the arithmetic series is the number of terms in the series times the average value of the terms. The average must be $\displaystyle (a_1+a_n)/2$ , since the values appear evenly spaced out around around this point on the real number line. Put another way, $\displaystyle (a_k+a_{n-k+1})/2, 1\leq k \leq n$ is constant and equal to $\displaystyle (a_1+a_n)/2$ , which corresponds to the fact that successively taking terms from opposite sides of the series gives a constant average, which therefore must be the average of all terms in the series.

#### Proof of the formula

Express the arithmetic series in two different ways:

$\displaystyle S_n=a_1+(a_1+d)+(a_1+2d)+\dots\dots+(a_1+(n-2)d)+(a_1+(n-1)d)$

$\displaystyle S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\dots\dots+(a_n-2d)+(a_n-d)+a_n$

Add both sides of the two equations. All terms involving d cancel, and so we're left with:

$\displaystyle \ 2S_n=n(a_1+a_n)$

Rearranging and remembering that $\displaystyle a_n = a_1 + (n-1)d$ , we get:

$\displaystyle S_n=\frac{n( a_1 + a_n)}{2}=\frac{n[ 2a_1 + (n-1)d]}{2}$ .

### Arithmetic series and sigma notation

Arithmetic series are commonly expressed using sigma notation. As an example, the arithmetic series

$\displaystyle a_1+(a_1+d)+(a_1+2d)+\dots\dots+(a_1+(n-2)d)+(a_1+(n-1)d) ,$

can be more succinctly written using sigma notation as

$\displaystyle \sum_{i=0}^{n-1} (a+id).$

Likewise, an arithmetic series

$\displaystyle a_1 + a_2 + a_3 + \dots\dots + a_{m-1} + a_m$

can be written as

$\displaystyle \sum_{j=1}^{m} a_j$

## Product

The product of the components of an arithmetic progression with an initial element $\displaystyle a_1$ , common distance $\displaystyle d$ , and $\displaystyle n$ elements in total, is determined in a closed expression by

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

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 $\displaystyle 1 \times 2 \times \ldots \times n$ is given by the factorial $\displaystyle n!$ and that the product

$\displaystyle m \times (m+1) \times \ldots \times (n-1) \times n \,\!$

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

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