The arithmetic series formula

The arithmetic series formula gives the sum of the first n terms of an arithmetic sequence: Sₙ = ½n[2a + (n − 1)d], where a is the first term and d is the common difference. If you know the last term l instead, use Sₙ = ½n(a + l): the number of terms times the average of the first and last. For example, the first 30 terms of 6, 10, 14, ... add up to ½ × 30 × (2 × 6 + 29 × 4) = 1920.

The arithmetic series formulas

The two versions are the same formula: substitute l = a + (n − 1)d into ½n(a + l) to get the first. AQA's AS and A-level formulae booklet gives both. The nth term itself is uₙ = a + (n − 1)d.

Proof of the arithmetic series formula

Pearson Edexcel's A-level specification says the proof of the sum formula should be known. Write the sum forwards and backwards:

Add the two lines term by term. Each pair adds to a + l, and there are n pairs:

Replacing l with a + (n − 1)d gives Sₙ = ½n[2a + (n − 1)d].

The sum of the first n natural numbers

With a = 1, d = 1 and l = n, the formula gives the sum 1 + 2 + 3 + ... + n:

Edexcel's specification mentions this formula alongside the proof. For example, 1 + 2 + ... + 100 = 100 × 101 ÷ 2 = 5050.

Arithmetic series formula examples

Example 1: from a, d and n

Add up the first 30 terms of 6, 10, 14, ...

Here a = 6, d = 4 and n = 30:

Example 2: from the first and last terms

Find 4 + 9 + 14 + ... + 99.

First find n: the terms go up in 5s, so 4 + (n − 1) × 5 = 99 gives n − 1 = 19 and n = 20. Then:

Example 3: finding the number of terms

An arithmetic series has a = 3 and d = 4. How many terms are needed for the sum to be 1176?

So 2n² + n − 1176 = 0. Using the quadratic formula, n = (−1 + √9409) ÷ 4 = (−1 + 97) ÷ 4 = 24. The negative solution is rejected, because n must be a positive whole number.

Example 4: two conditions

The 3rd term of an arithmetic series is 11, and the sum of the first 10 terms is 185. Find a and d.

From the 3rd term: a + 2d = 11. From the sum: 5(2a + 9d) = 185, so 2a + 9d = 37. Doubling the first equation gives 2a + 4d = 22, and subtracting, 5d = 15:

Example 5: sigma notation

Evaluate the sum from r = 1 to 25 of (4r − 1).

Writing out a few terms gives 3, 7, 11, ..., 99: an arithmetic series with a = 3, l = 99 and n = 25, so S = 12.5 × (3 + 99) = 1275. Alternatively, split it: 4 × (1 + 2 + ... + 25) − 25 = 4 × 325 − 25 = 1275.

Example 6: a savings plan

Sam saves £20 in the first month and £5 more each month than the month before. How much has Sam saved after 2 years?

24 months, a = 20, d = 5:

Sam has saved £1860.

Here is a quick check.

The Sequences and Series Calculator finds nth terms and sums of arithmetic and geometric series, and our arithmetic sequence calculator guide works through more nth-term problems.

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Arithmetic series vs geometric series

An arithmetic series adds the same amount each time; a geometric series multiplies by the same amount. Their sum formulas are different: the geometric version is Sₙ = a(1 − rⁿ) ÷ (1 − r). Check the pattern before choosing a formula: 3, 6, 9, 12 is arithmetic (add 3), but 3, 6, 12, 24 is geometric (multiply by 2).

Common mistakes

• Using n − 1 in the wrong place. It is (n − 1)d, inside the bracket, not ½(n − 1).
• Getting n wrong. Count the terms with l = a + (n − 1)d; the number of terms is not the last term.
• Mixing up uₙ and Sₙ. uₙ is one term; Sₙ is the sum of n terms.
• Keeping a negative or fractional n when solving the quadratic. n must be a positive whole number.
• Sigma limits. Σ from r = 1 to 20 has 20 terms; from r = 5 to 20 has 16 terms.

Practise sequences in a short timed paper

A-level Maths: Algebra & Functions gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so sequences and series turn up regularly but not in every paper.

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Maths · AlgebraArithmetic sequence calculator: nth term and sumFree arithmetic sequence calculator: the nth term rule, any term and the sum of n terms, from a and d or from the first few terms, with full working.