Arithmetic Sequence Formula: Find Any Term And The Sum
The short version
- The arithmetic sequence formula is an = a1 + (n – 1)d. Plug in the first term, the common difference and the term number to get any term.
- The common difference d is any term minus the term before it. It can be negative when the sequence counts down.
- To add up the first n terms, use Sn = n/2 × (a1 + an): the number of terms times the average of the first and last.
- The most common mistake is multiplying d by n instead of (n – 1). The first term already counts as one step.
The arithmetic sequence formula is an = a1 + (n – 1)d, where a1 is the first term, d is the common difference and n is the position of the term you want. For the sequence 5, 9, 13, 17…, the 20th term is 5 + 19 × 4 = 81.
That one arithmetic sequence formula answers most homework questions. Below we’ll also cover how to find which term a number is, how to build the formula from two terms, the recursive version, the sum formula, and practice problems with answers.
What makes a sequence arithmetic
A sequence is a list of numbers in order. It’s arithmetic when you add the same number every time to get the next term. That fixed number is the common difference, d.
- 3, 8, 13, 18… is arithmetic with d = 5.
- 40, 34, 28, 22… is arithmetic with d = -6.
- 2, 4, 8, 16… is not arithmetic. The gaps are 2, 4, 8, so they aren’t constant. (It’s geometric, because you multiply by 2 each time.)
To find d, subtract any term from the one right after it: second minus first, third minus second. If all the differences match, the sequence is arithmetic.
How the arithmetic sequence formula works
Think about how many jumps it takes to reach a term. To get from the 1st term to the 2nd, you add d once. To reach the 3rd, you add it twice. To reach the 20th term, you add it 19 times. That’s where n – 1 comes from.
an = a1 + (n – 1)d
| Symbol | Meaning | In 5, 9, 13, 17… |
|---|---|---|
| an | The term you’re looking for | Depends on n |
| a1 | The first term | 5 |
| d | Common difference | 4 |
| n | Position of the term | 1, 2, 3… |
Example 1: find a far-off term
Find the 20th term of 5, 9, 13, 17…
a1 = 5, d = 9 – 5 = 4, n = 20
a20 = 5 + (20 – 1) × 4
a20 = 5 + 19 × 4 = 5 + 76 = 81
You can also simplify the formula once and reuse it: an = 5 + 4(n – 1) = 4n + 1. Check it: n = 1 gives 5. Good.
Example 2: a sequence that counts down
Find the 10th term of 40, 34, 28…
d = 34 – 40 = -6
a10 = 40 + 9 × (-6) = 40 – 54 = -14
Example 3: which term is it?
In 5, 9, 13, 17…, which term equals 101? Set the formula equal to 101 and solve for n.
101 = 5 + (n – 1) × 4
96 = (n – 1) × 4
24 = n – 1
n = 25, so 101 is the 25th term.
If n comes out as a fraction, the number isn’t in the sequence at all.
Example 4: build the formula from two terms
The 3rd term is 11 and the 8th term is 31. Find the formula.
From term 3 to term 8 is 5 jumps, and the value grew by 31 – 11 = 20.
d = 20 ÷ 5 = 4
Back up two jumps from the 3rd term: a1 = 11 – 2 × 4 = 3
an = 3 + (n – 1) × 4, which simplifies to an = 4n – 1
Check: a8 = 32 – 1 = 31. It works.
Explicit vs. recursive formulas
The arithmetic sequence formula above is the explicit formula. It jumps straight to any term. The recursive formula tells you how to get each term from the one before it:
a1 = 5, an = an – 1 + 4
Recursive is easy to read but slow: to find the 100th term you’d need the 99th first. Most tests ask kids to switch between the two, so it helps to see that “+ 4” in the recursive version is the same d in the explicit one.
One more connection worth knowing: an arithmetic sequence is a linear function with the inputs limited to 1, 2, 3 and so on. In an = 4n + 1, the common difference 4 is the slope. Plot the terms and they land on a straight line.
| Arithmetic | Geometric | |
|---|---|---|
| Next term | Add d | Multiply by r |
| Example | 3, 7, 11, 15 | 3, 6, 12, 24 |
| Formula | an = a1 + (n – 1)d | an = a1 × rn – 1 |
| Graph | Straight line | Curve |
The sum formula for an arithmetic series
Adding up the terms of an arithmetic sequence gives an arithmetic series. The trick: pair the first and last terms, the second and second-to-last, and so on. Every pair has the same total. That gives:
Sn = n/2 × (a1 + an)
If you don’t know the last term, swap it out: Sn = n/2 × (2a1 + (n – 1)d).
Example 5: add 1 through 100
n = 100, a1 = 1, a100 = 100
S100 = 100/2 × (1 + 100) = 50 × 101 = 5,050
Example 6: seats in a theater
Row 1 has 20 seats and each row has 2 more than the row in front. There are 15 rows. How many seats in all?
Last row: a15 = 20 + 14 × 2 = 48
Total: S15 = 15/2 × (20 + 48) = 7.5 × 68 = 510 seats
Common mistakes
- Using n instead of n – 1. For the 20th term of 5, 9, 13…, writing 5 + 20 × 4 = 85 is off by one jump. The right answer is 81.
- Subtracting in the wrong order. d is the later term minus the earlier one. For 40, 34, 28, that’s -6, not 6.
- Dropping the negative sign. 40 + 9 × (-6) is 40 – 54. Kids who are shaky with negative numbers often write 94.
- Miscounting terms. How many terms in 7, 11, 15, …, 95? It’s (95 – 7) ÷ 4 + 1 = 23, not 22. Forgetting the “+ 1” is the counting version of the n – 1 mistake.
- Mixing up n and an. n is the position (5th, 20th). an is the value in that position.
Practice problems
- Find the common difference: 7, 12, 17, 22…
- Write the next three terms: 50, 43, 36…
- Find the 15th term of 3, 7, 11…
- Write an explicit formula for 10, 16, 22… and simplify it.
- Find the 12th term of 20, 17, 14…
- Which term of 2, 9, 16… equals 100?
- In an arithmetic sequence, a4 = 19 and a10 = 43. Find a1 and d.
- Find the sum of the first 30 terms of 4, 7, 10…
Answers
- d = 12 – 7 = 5
- 29, 22, 15 (d = -7)
- a15 = 3 + 14 × 4 = 59
- an = 10 + (n – 1) × 6 = 6n + 4
- a12 = 20 + 11 × (-3) = 20 – 33 = -13
- 100 = 2 + (n – 1) × 7, so 98 = 7(n – 1), n – 1 = 14, n = 15. It’s the 15th term.
- d = (43 – 19) ÷ (10 – 4) = 24 ÷ 6 = 4. a1 = 19 – 3 × 4 = 7.
- a30 = 4 + 29 × 3 = 91. S30 = 30/2 × (4 + 91) = 15 × 95 = 1,425.
If your kid is stuck on sequences
The arithmetic sequence formula has four pieces, and each one leans on an older skill. When a student can’t get sequence problems right, the idea of a pattern is usually fine. One of these older skills tends to be the weak spot:
- Adding and multiplying negative numbers (seventh grade), which breaks every decreasing sequence.
- Order of operations, so (n – 1) × d gets done before adding a1.
- Solving two-step equations, which is what “which term is 101?” really asks.
We wrote about this pattern in struggling with algebra: the trouble shows up in Algebra 1, but the cause is often years back. If you’re not sure whether it’s one topic or a wider gap, these signs by grade level help, and our guide to math assessment for kids explains how to find the missing piece without guessing.
Frequently asked questions
What is the formula for an arithmetic sequence?
an = a1 + (n – 1)d. a1 is the first term, d is the common difference and n is the term number. People sometimes search for the “arithmetic seq formula”; it’s this same one.
How do you find the common difference?
Subtract a term from the one after it. If you only have two terms that aren’t next to each other, divide the difference in values by the number of jumps between them.
How do you find the sum of an arithmetic sequence?
Use Sn = n/2 × (a1 + an). Multiply the number of terms by the average of the first and last terms.
Is an arithmetic sequence a linear function?
Yes, with whole-number inputs. The common difference is the slope. For example, an = 4n + 1 matches the line y = 4x + 1 at x = 1, 2, 3 and so on.
What’s the recursive formula for an arithmetic sequence?
Give the first term, then the rule: an = an – 1 + d. For 5, 9, 13…, that’s a1 = 5 and an = an – 1 + 4.
Can the common difference be zero or a fraction?
Yes to both. 6, 6, 6… has d = 0. 1, 1.5, 2, 2.5… has d = 0.5. The formula works the same way.
Find the actual gap
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