An arithmetic progression or arithmetic sequence is an ordered list of numbers such that the difference between the consecutive terms is constant. For instance, the sequence $5, 7, 9, 11, 13, \cdots$ is an arithmetic progression with common difference of $2$.

Finite Summation

The sum of the members of a finite arithmetic sequence is called an arithmetic series.

We can come up with a formula for the sum of a finite arithmetic formula by looking at the sum in two different ways. First we think of it as the sum of terms that are written in terms of $a_1$, so that the second term is $a_1+d$, the third is $a_1+2d$, and so on. Then our sum looks like:

$S_n = a_1 + \left(a_1 + d\right) + \left(a_1 +2d\right) + \cdots + \left[a_1 + \left(n-2\right)d\right] + \left[a_1 + \left(n-1\right)d\right]$

Next, we think of each term as being written in terms of the last term, $a_n$. Then the last term is $a_n$, the term before the last is $a_n-d$, the term before that is $a_n-2d$, and so on. The sum is therefore:

$S_n = \left[a_n - \left(n-1\right)d\right] + \left[a_n - \left(n-2\right)d\right] + \cdots + \left(a_n - 2d\right) + \left(a_n - d\right) + a_n$

Adding both sides of the two previous equations, all terms involving $d$ cancel and we obtain:

$2S_n = n\left(a_1 + a_n\right)$

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

$\displaystyle{S_n = \left( \frac{n}{2} \right) \left (a_1 + a_n \right )}$

This can be thought of as $n$ times the average of the first and last terms. An alternate form results from reinserting the substitution $a_n = a_1 + \left(n-1\right)d$:

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

For example, consider the arithmetic progression $3,8,13,18,23, \cdots$ and suppose you want to know the sum up to the $50$th term. Note that you will need:

• The number of terms you want to sum ($n$)
• The first term of the sequence ($a_1$)
• The difference in consecutive terms ($d$)

From the question, we know that $n=50$. We can see that the first term is $a_1 = 3$. The difference between the terms is $d = 5$. So the equation above gives: 

$\displaystyle{ \begin{aligned} S_{50} &=\left(\frac{50}{2}\right)\left(2 \cdot 3 + 49 \cdot 5\right) \\ &= 6275 \end{aligned} }$

Infinite Summation

An infinite arithmetic series is exactly what it sounds like: an infinite series whose terms are in an arithmetic sequence. Examples are $1 + 1 + 1 + 1 + \cdots$ and $1+2+3+4+ \cdots$. The general form for an infinite arithmetic series is:

$\displaystyle{\sum_{i=1}^\infty{a_i} = \sum_{i=1}^\infty{\left[a_1+\left(i-1\right)d\right]}}$

If $a_1=d=0$, then the sum of the series is $0$. If either $a_1$ or $d$ is non-zero, then the infinite series has no sum. 

Even if one is dealing with an infinite sequence, the sum of that sequence can still be found up to any $n$th term with the same equation used in a finite arithmetic sequence.