An arithmetic sequence is a series of numbers where the difference between successive numbers is constant throughout. The difference itself is known as the common difference.
\[1, 2, 3, 4, \dots\] \[3, 8, 13, 18, \dots\] \[17, 14, 11, 8, \dots\]Arithmetic sequences can also be thought of as linear functions, where the domain would be the term number in the sequence, and the range the actual number itself. The slope is just the common difference.
| $n$ | $a_n$ |
|---|---|
| 1 | 3 |
| 2 | 8 |
| 3 | 13 |
| 4 | 18 |
The variable $a$ with a subscript number is typically used to tell you which term is being referenced. So, in the sequence above, when $n=2$ then $a_2=8$.
One way to define an arithmetic sequence is with point-slope form, which produces an explicit definition. The output is clearly defined in terms of the input, like all other functions you’ve dealt with so far.
The sequence’s common difference, $d$, is actually the slope of the function, which gives us
\[y - a_n = d(x - n)\]And since you typically know the first term, you will see this more specific definition.
\[a_n - a_1 = d(n - 1) \quad \rightarrow \quad a_n = d(n - 1) + a_1\]Just keep in mind that it doesn’t have to be the first term in the sequence. If you happen to only know some other term, just swap in what you need. For example, if the third term in the sequence is 7 and the common difference is 4, then
\[a_n - 7 = 4(n - 3)\]There is also a recursive definition, meaning it is defined by the previous term in the sequence. The definition itself is made up of a base case, or starting point, and then defines the next term in the sequence.
\[a_n = \begin{cases} a_1 & n=1 \\ a_{n-1} + d & n > 1 \end{cases}\]Here’s the sequence above, where the first term is 3 and the common difference is 5.
\[a_n = \begin{cases} 3 & n=1 \\ a_{n-1} + 5 & n > 1 \end{cases}\]Given two terms, you can find the common difference if you remember that arithmetic sequences are just linear functions. If you are given $a_1=18$ and $a_5=26$, you technically know two points: $(1,18)$ and $(5,26)$. Use your definition of slope to find that the common difference is $8/4 = 2$.
What if the sequence is finite and you need to sum up all the terms? This is called an arithmetic series, and if the number of terms is small, you can brute force it and just add them up manually. But if it’s a large number of terms, there is a quicker way.
Let’s say we have a sequence, $S$, and we want to add up the first six terms in it, $S_6$. We’ll use the terms 2, 5, 8, 11, 14, 17. Because of the nature of arithmetic series, terms match up to produce the same sum. First goes with last, second goes with second to last, and so on. In this case, each pair adds up 19. We’ll take advantage of that fact to produce our equation.
\[\begin{align*} S_6 &= 2 + 5 + 8 + 11 + 14 + 17 \\ 2S_6 &= 19+19 +19 + 19 + 19 + 19 && \text{Sequence added to itself} \\ 2S_6 &= 6(19) \\ S_6 &= \frac{6(2 + 17)}{2} && \text{Split the 19 into first and last terms} \\ &= 57 \end{align*}\]Line 4 is the key here. Using just the number of terms, 6, and the first and last terms, 2 and 17, we can find the sum of an arithmetic sequence.
\[S_n = \frac{n(a_1 + a_n)}{2}\]What we did above is also called a summation, and the mathematical shorthand for them is sigma notation, which is the Greek letter $\Sigma$.
\[\sum_{i=1}^n a_i\]The index, $i$, tells you where you are starting, while $n$ tells you where to end, and $a_i$ is the $i$-th term in the sequence. So let’s take our summation from just above and rewrite it using summation notation.
First, we need the explicit definition. Let’s use the first term as our point, and the common difference is 3. And we’ll simplify it this time, too.
\[\begin{align*} a_n &= 3(n-1) + 2 \\ &= 3n - 1 \end{align*}\]Now we can rewrite our summation, with only the $n$ becoming an $i$ since we are using that as our index.
\[\sum_{i=1}^6 (3i - 1)\]Here’s another example of a summation.
\[\sum_{i=1}^9 (2i-6)\]Which translates to
\[\left(2\left(1\right)-6\right) + \left(2\left(2\right)-6\right) + \dots + \left(2\left(9\right)-6\right)\]Just like before, we only need to find the first and last terms, then we can use our equation from earlier to evaluate.
\[\begin{align*} \sum_{i=1}^9 (2i - 6) &= \frac{9\left(\left(2\left(1\right)-6\right)+\left(2\left(9\right)-6\right)\right)}{2} \\ &= \frac{9(-4+12)}{2} \\ &= \frac{72}{2} \\ &= 36 \\ \end{align*}\]