About the Arithmetic Sequence Calculator
An arithmetic sequence adds the same amount every step. 3, 8, 13, 18, 23 — always five more. That fixed amount is the common difference, and once you know it along with the starting value, every term in the sequence is determined forever.
It is the simplest interesting pattern in mathematics, and it describes a great deal of ordinary life: a fixed monthly saving, seats increasing by two per row in a theatre, a salary rising by the same amount each year, the depth of a well after each equal dig.
This calculator finds any term, sums any number of them, recovers the rule from two known terms, locates a value's position, and counts how many terms are needed to reach a running total.
How to Use the Arithmetic Sequence Calculator
From the first term and difference. The straightforward direction — lists the terms and gives their sum.
From two known terms. For when a question gives you the 3rd and 8th terms and expects you to work out the rule.
Which term is a given value? Answers "is 48 in this sequence, and where?"
How many terms to reach a total? For "how many months of saving before I have £2,000?"
Negative differences are fine, and so are decimals.
The Two Formulas
The nth term: aₙ = a₁ + (n − 1)d
The sum: Sₙ = n(a₁ + aₙ) / 2
= n[2a₁ + (n − 1)d] / 2Why it is n − 1, not n
This is the error to guard against above all others. The first term has had no difference added to it yet. Getting to the second takes one step, the third takes two, and the nth takes n − 1.
Using n instead overshoots by exactly one common difference every time — a small error that looks entirely plausible, which is what makes it dangerous.
Check it on the first term: a₁ = a₁ + (1 − 1)d = a₁. Correct. With n it would be a₁ + d, which is the second term.
Where the sum formula comes from
Write the series out forwards, then again backwards underneath it:
S = 3 + 8 + 13 + 18 + 23
S = 23 + 18 + 13 + 8 + 3
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2S = 26 + 26 + 26 + 26 + 26 = 5 × 26 = 130
S = 65Every column adds to the same total — the first term plus the last — and there are n columns. So 2S = n(a₁ + aₙ), and the sum is half of that.
The story attached to this is that Gauss, set to add the numbers from 1 to 100 as a schoolboy punishment, saw the pairing instantly: fifty pairs each totalling 101, giving 5,050. The story may be embellished but the method is genuinely this.
It also means the sum is just n × the average term, and since the terms are evenly spaced the average is simply the midpoint of the first and last.
Step-by-Step Example
Finding a term. The sequence starts at 3 and rises by 5. What is the 10th term?
a₁₀ = 3 + (10 − 1) × 5
= 3 + 9 × 5
= 3 + 45
= 48Nine steps, not ten. The sum of those ten terms:
S₁₀ = 10 × (3 + 48) / 2 = 10 × 25.5 = 255Recovering the rule. The 3rd term is 13 and the 8th is 38.
Five steps cover 38 − 13 = 25, so d = 25 / 5 = 5
Step back from the 3rd: a₁ = 13 − 2 × 5 = 3
Rule: aₙ = 5n − 2Two terms always suffice, however far apart, because the sequence has only two unknowns: where it starts and how fast it moves.
A value that is not there. Is 50 in the sequence 3, 8, 13, …?
n = (50 − 3) / 5 + 1 = 9.4 + 1 = 10.4Not a whole number, so no. It falls between the 10th term (48) and the 11th (53). A formula alone would hand you 10.4 as if it were an answer.
Saving to a target. £3 in month one, rising by £5 each month. When do you pass £200?
S₈ = 8 × [6 + 7 × 5] / 2 = 164
S₉ = 9 × [6 + 8 × 5] / 2 = 207Nine months.
Understanding Your Result
The result answers the specific question asked.
The rule appears in two forms: as aₙ = a₁ + (n − 1)d, and simplified to aₙ = dn + c. That second form makes it obvious that an arithmetic sequence is a straight line sampled at whole-number positions — the difference is the gradient.
The terms list the sequence, trailing off with the final term shown if the list is long.
The sum gives the total and the mean term.
The notes describe the sequence's direction and character.
Two answers a formula alone cannot give
Whether a value appears at all. Positions must be whole numbers, so most values simply are not in the sequence. The calculator says which two terms a near miss falls between.
How many terms reach a total. This is counted rather than solved. Inverting the sum formula gives a quadratic with two roots, and with a negative difference the running total can rise, peak, and fall back past the target — so the algebraic root is not always the answer anyone wants. Counting gives the first n that genuinely reaches it.
When Should You Use This Calculator?
Savings and repayments. A fixed amount added each period, and the running total.
Seating and stadiums. Rows increasing by a fixed number of seats — the total capacity is an arithmetic sum.
Depreciation, straight-line. An asset losing the same value each year.
Stacking and stock. Logs, pipes or tins stacked in rows that shrink by one.
Salary progression. Fixed annual increments, and lifetime earnings as the sum.
Coursework. The nth term and the sum are standard exercises, and the working shows where a hand attempt diverged.
Any evenly spaced measurement. Readings at regular intervals, marks along a ruler, the positions of fence posts.
Common Mistakes
Using n instead of n − 1. Overshoots by one difference. The commonest error in the topic.
Forgetting to halve in the sum formula. n(a₁ + aₙ) is twice the sum.
Confusing the sequence with the series. The sequence is the list of terms; the series is their total. A question asking for "the sum of the first 20 terms" wants the series.
Treating a geometric sequence as arithmetic. 2, 6, 18, 54 is multiplying by 3, not adding a fixed amount. Check the differences between consecutive terms are constant before applying any of this.
Assuming a value must be in the sequence. Most are not. The position has to come out as a whole number.
Mishandling a negative difference. All the formulas work unchanged. Adding −3 is subtracting 3, and the terms will eventually pass through zero and turn negative.
Counting terms wrongly in a range. The terms from the 5th to the 12th number eight, not seven: 12 − 5 + 1. Off-by-one errors here are common.
Rounding the difference too early. If d comes out as 1/3, keep it as a fraction through the working rather than 0.33.