Arithmetic Sequences and Series
Stack a wall of bricks so each row holds two more bricks than the row below it. Row 1 has 5 bricks, row 2 has 7, row 3 has 9. Each row adds the same amount, so the row counts form an arithmetic sequence, and the total brick count is an arithmetic series. The same pattern shows up in seats per row in a theater, savings that grow by a fixed deposit, and stacked-pipe problems.
By the end of this page you can:
- Test whether a sequence is arithmetic by checking for a constant difference between consecutive terms.
- Write the nth term of an arithmetic sequence from its first term and common difference.
- Add the first n terms of an arithmetic sequence using the partial-sum formula.
The floor. An arithmetic sequence adds the same number, the common difference $d$, to get each new term, so $a_n = a_1 + (n-1)d$. The sum of the first $n$ terms is $n$ times the average of the first and last term.
Quick Reference
| Field | Value |
|---|---|
| Textbook | Stewart, Redlin, Watson 7e |
| Chapter | 12 |
| Section | 12.2 Arithmetic Sequences |
| Open alternate | OpenStax Precalculus 2e, 11.2 Arithmetic Sequences |
| Difficulty | Core |
| Time | About 30 minutes |
Before You Start
The Idea
A sequence is arithmetic when the gap between consecutive terms never changes. That fixed gap is the common difference $d$, found by subtracting any term from the one after it: $d = a_{n} - a_{n-1}$. Once you know the first term $a_1$ and the common difference $d$, every term is locked in, because you start at $a_1$ and step up by $d$ a total of $n - 1$ times to reach $a_n$.
To add up many terms at once, pair the first term with the last, the second with the second-to-last, and so on. Each pair has the same total, $a_1 + a_n$. That trick gives the partial sum: the number of terms $n$ times the average of the first and last term.
Key Formulas
The nth term of an arithmetic sequence with first term $a_1$ and common difference $d$:
\[ a_n = a_1 + (n - 1)\,d \]
Source: Stewart 7e, Section 12.2 Arithmetic Sequences.
The sum of the first $n$ terms (the nth partial sum):
\[ S_n = \frac{n\,(a_1 + a_n)}{2} = \frac{n}{2}\,\bigl(2a_1 + (n - 1)\,d\bigr) \]
Source: Stewart 7e, Section 12.2 Arithmetic Sequences.
Worked Example
A theater has 20 seats in the first row, and each row behind it has 3 more seats than the row in front. How many seats are in row 10, and how many seats are in the first 10 rows total?
Identify the parts. The first term is $a_1 = 20$ and the common difference is $d = 3$, since each row adds 3 seats. You want $a_{10}$ and $S_{10}$.
Show the solution
Find the 10th term:
\[ a_{10} = a_1 + (10 - 1)\,d = 20 + 9 \cdot 3 = 20 + 27 = 47. \]
Row 10 has 47 seats.
Find the sum of the first 10 terms:
\[ S_{10} = \frac{10\,(a_1 + a_{10})}{2} = \frac{10\,(20 + 47)}{2} = \frac{10 \cdot 67}{2} = \frac{670}{2} = 335. \]
The first 10 rows hold 335 seats.
Common slips to avoid: there are 9 steps of size $d$ to reach the 10th term, not 10, so the count of terms in the sum is $n = 10$, not 11. The common difference is 3, the size of each step, not the starting value 20.
Connections
This builds on sequences and summation notation, where you first met the nth term $a_n$ and the idea of adding a finite list of terms. It sits beside geometric sequences, which multiply by a fixed ratio instead of adding a fixed difference, and the two are the standard contrast pair. The partial-sum formula here is a clean target for mathematical induction, which proves such a formula holds for every $n$. The pairing argument behind the sum is the same idea that gives the formula for the sum of the first $n$ whole numbers.
Resources
- Stewart, Redlin, Watson 7e, Section 12.2 Arithmetic Sequences (primary).
- OpenStax Precalculus 2e, Section 11.2 Arithmetic Sequences (open alternate).
Check your understanding
An arithmetic sequence has first term 4 and common difference 5. What is the 8th term?
Last updated: 2026-06-24