Arithmetic Sequence and Series Calculator for Terms, Common Difference and Finite Sums

Get the nth term rule and any term or sum, rebuild a sequence from two known terms, find where a value sits, or paste a list to see what kind of sequence it is.

Choose what to calculate

Values can be whole numbers, decimals or fractions such as 3/4 or -2/5.

Arithmetic sequence and series formulas

In an arithmetic sequence you get from one term to the next by adding the same number every time. That number is the common difference, d. Add the terms up and you have an arithmetic series.

aₙ = a₁ + (n − 1)d
aₙ = dn + (a₁ − d)
aₙ = aₙ₋₁ + d
Sₙ = n(a₁ + aₙ) ÷ 2
Sₙ = n[2a₁ + (n − 1)d] ÷ 2

The first two lines are the same rule written two ways. The second is the simplified form most exams want: for 3, 7, 11, 15, … that’s 4n − 1, because d = 4 and 3 − 4 = −1. The third line is the recursive version, which says how to get each term from the one before but can’t jump straight to term 50. The calculator shows the simplified rule with every result.

Finding a sequence from two known terms

If you know any two terms and where they sit, divide the change in value by the change in position:

d = (aₖ₂ − aₖ₁) ÷ (k₂ − k₁)
a₁ = aₖ − (k − 1)d

Say a₂ = 7 and a₅ = 19. Then d = (19 − 7) ÷ (5 − 2) = 4, and a₁ = 7 − 4 = 3. One of the two can be the first term itself, so this mode also finds d when you know a₁ and any later term.

Finding where a value sits

Rearrange the nth-term formula to find a value’s position: n = (aₙ − a₁) ÷ d + 1. A positive whole-number result means the value is a term in the sequence. Otherwise, it lies between two terms, which the calculator identifies. You can use the same calculation to count the terms in 12, 19, 26, …, 397 by treating 397 as aₙ.

Adding a range of terms

To add terms m to n, take the sum up to term n and remove everything before term m:

Sum from m to n = Sₙ − Sₘ₋₁

Sₘ₋₁ holds exactly the earlier terms you don’t want, so subtracting it leaves just your range.

Worked example and how sequences behave

Take 3, 7, 11, 15, …, which has a first term of 3 and a common difference of 4. The 10th term is 3 + 9 × 4 = 39, and the first 10 terms add up to 10 × (3 + 39) ÷ 2 = 210.

The sign of d sets the direction. Add a positive number and the terms climb, add a negative one and they fall, and with d = 0 the same number just repeats. Plot the terms against their positions and you always get points on a straight line, which is why the nth term rule looks like the equation of a line.

You’ll spot arithmetic patterns in a taxi fare that adds the same charge for every extra mile, a staircase where each step rises by the same height, a training plan that adds five minutes each week, or a loan repaid in equal amounts with no interest, where the balance falls by the same sum every month.

Unlike some geometric series, an arithmetic series has no sum to infinity. Keep adding and the total grows or falls without limit, unless every term is zero.

Arithmetic, geometric and quadratic sequences

Arithmetic sequences add the same amount each step: 2, 5, 8, 11 adds 3. Geometric sequences multiply by the same amount: 2, 6, 18, 54 multiplies by 3. Quadratic sequences have differences that change by the same amount: in 2, 5, 10, 17 the differences 3, 5, 7 go up by 2 each time. Paste any list into the check mode and the calculator tells you which type it is.

Common mistakes

The most common slip is forgetting the “− 1” in a₁ + (n − 1)d. When n = 1, the multiplier n − 1 is zero; consequently, term 10 is nine equal steps after the starting term. Another mistake is finding d in the wrong direction. Subtract each term from the term after it, so 20, 17, 14 has d = −3, not 3. When simplifying a rule, students may also use the first term as the constant and write 4n + 3 for 3, 7, 11. The constant is actually a₁ − d, giving 4n − 1. Some books number the starting term as term 0; the final question below explains how to translate between the two conventions.

Arithmetic sequence practice questions

These are in the style of GCSE and A-level sequence questions. The numbers are our own, and you can check each answer with the calculator.

What is the nth term of 5, 11, 17, 23, …?

6n − 1. The terms go up by 6, so the rule starts 6n. The first term is 5, and 6 × 1 = 6, so you need to take off 1. Check: 6 × 4 − 1 = 23.

Is 100 a term in the sequence 7, 11, 15, 19, …?

No. The nth term is 4n + 3. Setting 4n + 3 = 100 gives n = 24.25, which isn’t a whole number. 100 falls between term 24 (99) and term 25 (103).

How many terms are in 12, 19, 26, …, 397?

56. The difference is 7, so the position of 397 is (397 − 12) ÷ 7 + 1 = 55 + 1 = 56. A common mistake is to forget the + 1 and answer 55.

What is the sum of the first 30 terms of 2, 9, 16, …?

3,105. The 30th term is 2 + 29 × 7 = 205, so the sum is 30 × (2 + 205) ÷ 2 = 15 × 207 = 3,105.

The 4th term is 10 and the 9th term is 35. What are the first term and the nth term?

The value rises by 25 over 5 steps, so d = 5. Going back three steps from the 4th term gives a₁ = 10 − 15 = −5. The nth term is 5n − 10.

How many terms of 5, 9, 13, … do you need for the sum to go over 500?

16. The sum of n terms is n(2n + 3), so you need 2n² + 3n > 500. Fifteen terms give 495, which falls just short, and sixteen give 560. In the calculator, set n to 15 and then 16 and compare the sums.

Can I enter fractions?

Yes. Enter a simple fraction such as 1/3 or -5/2. This is more accurate than shortening one third to 0.333 before the calculation, although very long results may still be rounded for display.

My textbook starts at term 0. How do I use the calculator?

If the book calls its starting value a₀, enter that value as the calculator’s first term. The calculator’s a₁ then represents the book’s a₀, so calculator term aₙ corresponds to book term aₙ₋₁. For example, calculator term a₁₀ is the book’s a₉. If you instead enter the book’s a₁ = a₀ + d, the term numbers match from a₁ onward.

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