An arithmetic sequence adds the same number every time: $7, 10, 13, 16, \ldots$ adds $3$. That number is the common difference $d$. Since a sequence is a function of position, an arithmetic sequence is a linear function in disguise, like the lines $y = mx + b$ you know, with slope $d$.
Why it works
To get from the first term to the $n$th term, how many times do you add $d$? From term $1$ to term $2$ is one step, to term $3$ is two steps, so to term $n$ it is $n - 1$ steps. That is where the formula comes from: $a_n = a_1 + (n - 1)d$. Expanding it, $a_n = dn + (a_1 - d)$, a line with slope $d$, evaluated only at whole-number positions.
Key idea
$a_n = a_1 + (n - 1)d$: start at the first term and add the common difference $n - 1$ times.
Worked examples
Worked example: A far-away term
Find the $40$th term of $7, 10, 13, 16, \ldots$
- $a_1 = 7$ and $d = 10 - 7 = 3$.
- $a_{40} = 7 + (40 - 1) \cdot 3 = 7 + 117 = 124$.
Answer: $124$
Worked example: A formula
Write a formula for $a_n$ in $20, 16, 12, 8, \ldots$
- $a_1 = 20$ and $d = 16 - 20 = -4$.
- $a_n = 20 + (n - 1)(-4) = 20 - 4n + 4 = -4n + 24$.
- Check: $n = 1$ gives $20$ and $n = 2$ gives $16$.
Answer: $a_n = -4n + 24$
Worked example: Two known terms
An arithmetic sequence has $a_3 = 11$ and $a_8 = 31$. Find $d$ and $a_1$.
- From term $3$ to term $8$ is $5$ steps: $5d = 31 - 11 = 20$, so $d = 4$.
- Back from term $3$ to term $1$ is $2$ steps: $a_1 = 11 - 2(4) = 3$.
Answer: $d = 4$, $a_1 = 3$.
Common mistake
Writing $a_n = a_1 + nd$. That adds $d$ one time too many: for $n = 1$ it gives $a_1 + d$, the second term. Check any formula by putting in $n = 1$; you must get the first term back.