Math Core

When a quantity changes by the same amount at every step, such as seats added to each row of a theater or a fixed deposit each week, its values form an arithmetic sequence. In this lesson you'll write formulas for these sequences and then find their sums quickly, even when there are hundreds of terms.

Common difference

In the sequence $7, 11, 15, 19, 23, \dots$ you add $4$ each time. Subtracting any term from the next one always gives the same number.

Definition

Arithmetic sequence

A sequence is arithmetic if the difference between consecutive terms is constant. That constant $d = a_n - a_{n-1}$ is the common difference.

The common difference can be positive (terms increase), negative (terms decrease) or zero (every term is the same). For $20, 14, 8, 2, \dots$ , $d = 14 - 20 = -6$ .

The explicit formula

To reach the $n$ th term, start at $a_1$ and add $d$ one time fewer than the position number:

a_2 = a_1 + d, \qquad a_3 = a_1 + 2d, \qquad a_4 = a_1 + 3d, \qquad \dots

nth term of an arithmetic sequence

a_n = a_1 + (n - 1)d

Distributing gives a linear expression in $n$ . For $7, 11, 15, \dots$ the formula is $a_n = 7 + 4(n - 1) = 4n + 3$ . That's no coincidence: an arithmetic sequence is a linear function restricted to the positive integers. The common difference is the slope, and plotting the terms gives points on a line.

The terms 7, 11, 15, 19, 23 lie on the line y = 4x + 3, whose slope is the common difference 4.Open in grapher →

Worked example: A far-off term

An arithmetic sequence has $a_1 = 7$ and $d = -3$ . Write an explicit formula and find $a_{20}$ .

a_n = 7 + (n - 1)(-3) = 7 - 3n + 3 = 10 - 3n.

Then $a_{20} = 10 - 3(20) = -50$ .

Working from two terms

If you know two terms but not the start, use the fact that moving forward $k$ positions adds $kd$ .

Worked example: Two terms known

In an arithmetic sequence, $a_4 = 19$ and $a_{11} = 47$ . Find an explicit formula.

From position $4$ to position $11$ is $7$ steps, so

47 = 19 + 7d \quad\Rightarrow\quad 7d = 28 \quad\Rightarrow\quad d = 4.

Back up $3$ steps from $a_4$ : $a_1 = 19 - 3(4) = 7$ . So $a_n = 7 + 4(n - 1) = 4n + 3$ .

Check: $a_{11} = 44 + 3 = 47$ . ✓

You can also ask which term has a given value. In $7, 11, 15, \dots$ , which term is $203$ ? Solve $4n + 3 = 203$ to get $n = 50$ . If the solution isn't a positive integer, the number isn't in the sequence.

Arithmetic series

An arithmetic series is the sum of the terms of an arithmetic sequence. Here's a neat trick for finding it. Write the sum $S = 1 + 2 + 3 + \cdots + 100$ forward and backward, then add the two lines:

\begin{array}{rcccccccc} S &=& 1 &+& 2 &+& \cdots &+& 100 \\ S &=& 100 &+& 99 &+& \cdots &+& 1 \\ \hline 2S &=& 101 &+& 101 &+& \cdots &+& 101 \end{array}

Every column adds to $101$ , and there are $100$ columns, so $2S = 100 \cdot 101$ and $S = 5050$ .

The same trick works for any arithmetic series. Pairing the first and last terms, the second and second-to-last, and so on, always gives the same total $a_1 + a_n$ , because as one term goes up by $d$ its partner goes down by $d$ .

Sum of an arithmetic series

The sum of the first $n$ terms is

S_n = \frac{n(a_1 + a_n)}{2} \qquad\text{or, equivalently,}\qquad S_n = \frac{n}{2}\bigl[2a_1 + (n - 1)d\bigr].

In words: the number of terms times the average of the first and last terms.

Use the first form when you know the last term, and the second when you know $d$ but not the last term.

Common mistake

The most common error is miscounting $n$ . The number of terms from $a_1$ to $a_n$ is $\dfrac{a_n - a_1}{d} + 1$ . For $5, 9, 13, \dots, 101$ , that's $\dfrac{96}{4} + 1 = 25$ terms, not $24$ . Forgetting the $+1$ is a classic "fence-post" mistake.

Worked example: Summing a listed series

Find $5 + 9 + 13 + \cdots + 101$ .

Here $a_1 = 5$ , $d = 4$ and the last term is $101$ . As computed in the warning, there are $n = 25$ terms. So

S_{25} = \frac{25(5 + 101)}{2} = \frac{25 \cdot 106}{2} = 1325.

Worked example: A series in sigma notation and a word problem

(a) Evaluate $\displaystyle\sum_{k=1}^{30} (4k - 1)$ .

The terms are linear in $k$ , so the series is arithmetic. The first term is $4(1) - 1 = 3$ and the last is $4(30) - 1 = 119$ . There are $30$ terms:

S_{30} = \frac{30(3 + 119)}{2} = 15 \cdot 122 = 1830.

(b) A theater has $15$ rows. The first row has $20$ seats, and each row has $3$ more seats than the row in front of it. How many seats are there in all?

The seat counts form an arithmetic sequence with $a_1 = 20$ , $d = 3$ . The last row has $a_{15} = 20 + 14(3) = 62$ seats. The total is

S_{15} = \frac{15(20 + 62)}{2} = \frac{15 \cdot 82}{2} = 615 \text{ seats}.

Tip

Any sum $\sum (mk + b)$ with a linear expression in $k$ is an arithmetic series with common difference $m$ . Find the first and last terms by plugging in the limits, count the terms, and use $S_n = \dfrac{n(a_1 + a_n)}{2}$ .