Arithmetic Sequence Formula Explained (With Examples)
By ToolNimba Editorial Team June 20, 2026 7 min read
Quick answer
The nth term of an arithmetic sequence is a_n = a_1 + (n - 1) x d, where a_1 is the first term, n is the position of the term you want, and d is the common difference (the fixed amount added between terms). To find the sum of the first n terms, use S_n = n / 2 x (a_1 + a_n).
An arithmetic sequence is a list of numbers where each term changes by the same fixed amount. That amount is called the common difference. Because the step never changes, you do not have to write out every term to find a far-off one. A single formula jumps straight to any position you want, whether it is the 5th term or the 500th.
This guide breaks the arithmetic sequence formula into plain language, walks through worked examples step by step, and shows you how to add up a whole series quickly. If you are revising other math foundations, the distance formula explained and slope formula explained posts pair nicely with this one.
What is an arithmetic sequence?
An arithmetic sequence (sometimes called an arithmetic progression) is any ordered list of numbers in which the gap between one term and the next stays constant. You build it by repeatedly adding the same number. That number can be positive, negative, or even a fraction.
Here are a few quick examples to make it concrete:
- 3, 7, 11, 15, 19, ... adds 4 each time, so the common difference d is 4.
- 20, 17, 14, 11, ... subtracts 3 each time, so d is -3 (a decreasing sequence).
- 1, 1.5, 2, 2.5, ... adds 0.5 each time, so d is 0.5.
- 100, 100, 100, ... has d equal to 0, which is a valid but constant sequence.
To check whether a list is arithmetic, subtract any term from the one after it. If you get the same answer every time, it is arithmetic. If the difference changes, it is not. A sequence that multiplies by a fixed factor instead of adding is a different beast covered in the geometric sequence formula explained guide.
The arithmetic sequence formula for the nth term
The core formula lets you find any term without listing the ones before it:
The nth term formula
a_n = a_1 + (n - 1) x d
Each piece has a clear job:
What each symbol in the formula means
| Symbol | Meaning | Example |
|---|---|---|
| a_n | The term you are solving for, at position n | the 10th term |
| a_1 | The first term in the sequence | 3 |
| n | The position of the term you want | 10 |
| d | The common difference between consecutive terms | 4 |
The reason it works is simple. To reach the nth term from the first term, you take exactly n minus 1 steps, and each step adds d. So you add d a total of (n - 1) times. That is why the formula uses (n - 1) and not n. Forgetting that single subtraction is the most common error people make, and we cover it again in the mistakes section below.
Worked example: find the 12th term
Suppose the sequence is 5, 9, 13, 17, ... and you want the 12th term. Follow these steps:
- Identify the first term: a_1 = 5.
- Find the common difference by subtracting consecutive terms: 9 - 5 = 4, so d = 4.
- Note the position you want: n = 12.
- Plug into the formula: a_12 = 5 + (12 - 1) x 4.
- Simplify the bracket: a_12 = 5 + 11 x 4.
- Multiply, then add: a_12 = 5 + 44 = 49.
So the 12th term is 49. Notice you never had to write out terms 5 through 11. The formula skipped straight to the answer, which is exactly why it is so useful for large positions.
Finding the common difference or the first term
The same formula can be rearranged to solve for whatever piece is missing. If you know two terms, you can back out d, and from there the first term.
Solving for the common difference
If you know two terms and their positions, the common difference is the change in value divided by the change in position. For terms a_m and a_n, d = (a_n - a_m) / (n - m). For example, if the 3rd term is 11 and the 8th term is 31, then d = (31 - 11) / (8 - 3) = 20 / 5 = 4.
Solving for the first term
Once you have d, rearrange the main formula to a_1 = a_n - (n - 1) x d. Using the example above with the 3rd term equal to 11 and d = 4: a_1 = 11 - (3 - 1) x 4 = 11 - 8 = 3. The sequence therefore starts at 3.
Explicit vs recursive formula
There are two standard ways to describe an arithmetic sequence, and exams often ask you to switch between them. The explicit formula is the one we have used so far, a_n = a_1 + (n - 1) x d. It jumps straight to any term using only its position, so it is the right tool when you need a far-off term such as the 100th.
The recursive formula instead defines each term using the one before it: a_1 is given, and a_n = a_(n-1) + d. It mirrors how the sequence is actually built, one step at a time, but it is slow for large positions because you must climb through every earlier term first. The table below contrasts the two.
Explicit and recursive forms compared
| Form | Definition | Best for |
|---|---|---|
| Explicit | a_n = a_1 + (n - 1) x d | Finding a single distant term quickly |
| Recursive | a_1 given, a_n = a_(n-1) + d | Showing the step by step build of the list |
To convert a recursive rule into an explicit one, read off a_1 and d, then drop them into a_n = a_1 + (n - 1) x d. To go the other way, state the first term and add that the next term is the previous term plus d. Both descriptions produce the exact same numbers.
Adding up the terms: the sum formula
When you add the terms of an arithmetic sequence, you get an arithmetic series. There are two equivalent formulas for the sum of the first n terms, and you pick whichever matches the information you have.
Two ways to find the sum of n terms
| Use this when you know | Formula |
|---|---|
| The first and last terms | S_n = n / 2 x (a_1 + a_n) |
| The first term and the common difference | S_n = n / 2 x (2 x a_1 + (n - 1) x d) |
Here is the same calculation laid out as steps. Add the first 10 terms of 5, 9, 13, ...
- Identify the first term: a_1 = 5 and the common difference d = 4.
- Find the last term you need with the nth term formula: a_10 = 5 + (10 - 1) x 4 = 5 + 36 = 41.
- Drop the values into the sum formula: S_10 = 10 / 2 x (5 + 41).
- Simplify: S_10 = 5 x 46 = 230.
Why does halving n and multiplying by the first plus last term work? Pair the first term with the last, the second with the second to last, and so on. Every pair adds up to the same total (a_1 + a_n), and there are n / 2 such pairs. Multiply the pair total by the number of pairs and you have the sum. This is the pairing trick the young mathematician Carl Friedrich Gauss reportedly used to add 1 through 100 in seconds. If you only need a fast running total of plain numbers, the average calculator and a quick multiplication often get you there even faster.
Common mistakes to avoid
- Using n instead of (n - 1). You only add the common difference n minus 1 times, because the first term needs zero steps. Always subtract 1.
- Getting the sign of d wrong. In a decreasing sequence d is negative. Subtract terms in the order later minus earlier, not the reverse.
- Assuming every sequence is arithmetic. Check that the difference is truly constant before applying the formula. If it changes, the sequence may be geometric or neither.
- Confusing the term value with its position. a_n is the value at position n, not n itself. Mixing these up leads to wrong answers fast.
- Forgetting order of operations. Do the bracket and multiplication before the final addition, just like in the worked steps above.
Arithmetic sequence formulas at a glance
Keep this reference handy. Each row tells you which formula to reach for based on what you are trying to find.
Quick reference for every arithmetic sequence formula
| Goal | Formula |
|---|---|
| Find the nth term (explicit) | a_n = a_1 + (n - 1) x d |
| Define each term (recursive) | a_n = a_(n-1) + d, with a_1 given |
| Find the common difference | d = (a_n - a_m) / (n - m) |
| Find the first term | a_1 = a_n - (n - 1) x d |
| Sum with first and last term | S_n = n / 2 x (a_1 + a_n) |
| Sum with first term and d | S_n = n / 2 x (2 x a_1 + (n - 1) x d) |
Where arithmetic sequences show up
These sequences are not just textbook exercises. Anything that grows or shrinks by a fixed amount per step fits the pattern: saving the same amount each month, seats that increase by a set number per stadium row, simple interest that adds a fixed sum per period, or a runner adding the same distance to each training week. For percentage based growth instead, see how to calculate percentage or our percentage calculator.
If you need to average a set of sequence terms or any other numbers, the average calculator handles it instantly, and the how to calculate average guide explains the method behind it.