Consider a sequence of numbers where the first term is $a_1$, the second term is $a_2$, the third term is $a_3$, and so on. The common difference, or constant that is being added to each term is $d$, and $a_n$ is the $n^{\rm th}$ term.

A pattern (formula) emerges. Using the previous information, we obtain:

$a_1=a_1$
$a_2 = a_1 + d$
$a_3 = a_2 + d = a_1 + d + d = a_1 + 2d$
$a_4 = a_3 + d = a_1 + 2d + d = a_1 + 3d$
$a_5 = a_4 + d = a_1 + 3d + d = a_1 + 4d$
$\vdots$
$a_n = a_1 + (n − 1)d$      

This is the formula for the $n^{\rm th}$ term of an arithmetic sequence.

Example 1: Consider the sequence of numbers $5$, $9$, $13$, $17$, $21 \ldots$
Find the $\rm 100^{th}$ term. That is to say, find $a_{100}$.

We know $a_1 = 5$, $d = 4$, and $n = 100$ since we are looking for the $\rm 100^{th}$ term. Using the formula for the $n^{\rm th}$ term of an arithmetic sequence:

$a_n = a_1 + (n − 1)d$
$a_{100} = 5 + (100 − 1)(4)$
$a_{100} = 5 + (99)(4)$
$a_{100} = 5 + 396$
$a_{100} = 401$

Thus the $\rm 100^{th}$ term in this sequence will be $401$.

If we wanted to determine the sum of all the terms, then another pattern (formula) emerges. Without going into a formal proof, the sum of the first $n$ terms, $\mathrm S_n$, of an arithmetic sequence, called an arithmetic series, is given by:

$\mathrm S_n = \dfrac{n}{2}(1a_1 + (n-1)d)$.

Therefore, we need to know a few pieces of information. Namely, the first term $a_1$, how many terms are we summing $n$, and the common difference $d$.

Notice that $2a_1 = a_1 + a_1$. Thus, this can also be written as:

$\mathrm S_n = \dfrac{n}{2}(a_1 + a_1 + (n-1)d)$
$\mathrm S_n = \dfrac{n}{2}(a_1 + a_n)$

Therefore, we need to know a few different pieces of information. Namely, the first term $a_1$, how many terms are we summing $n$, and the last term in the sequence $a_n$.

Example 2: Consider the previous sequence of numbers $5$, $9$, $13$, $17$, $21 \ldots$
Determine $\mathrm S_{100}$.

The sum of the first $100$ terms, $\mathrm S_{100}$, can be computed as:

$\mathrm S_{100} = \dfrac{100}{2}(2(5) + (100-1)(4)$
$\mathrm S_{100} = 50(10 + (99)(4))$
$\mathrm S_{100} = 50(10 + 396)$
$\mathrm S_{100} = 50(406)$
$\mathrm S_{100} =20~300$