How to Find the Sum of an Arithmetic Sequence: IGCSE Maths
The sum of an arithmetic sequence is what you get when you add up all the terms of a linear sequence: a few terms, the first hundred, or any number you choose. Rather than adding them one by one, a single formula does the work in one step. This page covers the formula Sn = (n/2)[2a + (n-1)d], where it comes from, and how to apply it to typical IGCSE problems including word problems with savings, seating, and stacked rows. It is part of the Sequences chapter of the Edexcel IGCSE Maths course.
What is an arithmetic series?
If you take an arithmetic sequence and add up the terms, the result is called an arithmetic series. For example, the sequence \(3, 7, 11, 15, 19, \ldots\) becomes the series \(3 + 7 + 11 + 15 + 19 + \ldots\) when you add the terms together.
The notation \(S_n\) means the sum of the first n terms. So \(S_5\) is the sum of the first five terms, \(S_{20}\) is the sum of the first twenty, and so on.
Adding term by term works fine for short series, but slow and error-prone for long ones. A single formula gives the answer in one step.
Show the algebraic proof
The same pairing trick generalises to any arithmetic sequence with first term \(a\) and common difference \(d\). Write the sum forwards, then directly underneath write it backwards. Each column then adds to the same total.
Notice how the \(d\) terms cancel column by column: in the first column \(a\) plus \(a+(n-1)d\) gives \(2a+(n-1)d\); in the second column the extra \(+d\) on the top is matched by an extra \(-d\) on the bottom, so the total is again \(2a+(n-1)d\). Every column gives the same value.
The two rows together add up to \(2S_n\), and there are \(n\) columns each contributing \(2a+(n-1)d\):
Divide both sides by \(2\):
The arithmetic series formula
To use the formula you need three pieces of information:
- \(a\): the first term
- \(d\): the common difference
- \(n\): how many terms you want to add
Substitute these in and the formula returns \(S_n\), the total. The square brackets around \(2a + (n-1)d\) act like a single number; calculate it first, then multiply by \(\tfrac{n}{2}\) at the end.
Three-step method
- Identify the first term \(a\) and common difference \(d\) from the sequence.
- Decide how many terms \(n\) you are summing.
- Substitute into \(S_n = \tfrac{n}{2}[2a + (n-1)d]\) and simplify.
Worked examples
💡 Example 1: direct application
Find the sum of the first 20 terms of the sequence \(5, 8, 11, 14, \ldots\)
\(a = 5\), \(d = 3\), \(n = 20\)
\(S_{20} = \dfrac{20}{2}\bigl[2(5) + (20-1)(3)\bigr]\)
\(= 10[10 + 57]\)
\(= 10 \times 67\)
\(= 670\)
What's happening?
Pull \(a\) and \(d\) straight from the sequence. Substitute, simplify the bracket first, then multiply.
💡 Example 2: find n given the sum
For the sequence \(2, 5, 8, 11, \ldots\), how many terms are needed for the sum to equal 222?
\(a = 2\), \(d = 3\), \(S_n = 222\)
\(222 = \dfrac{n}{2}\bigl[2(2) + (n-1)(3)\bigr]\)
\(444 = n[4 + 3n - 3]\)
\(444 = n[3n + 1]\)
\(3n^2 + n - 444 = 0\)
Factorising: \((3n + 37)(n - 12) = 0\)
So \(n = 12\). The negative root is rejected.
What's happening?
You set \(S_n\) equal to the target sum and solve for \(n\). The result is a quadratic. Reject any non-positive or non-integer roots since \(n\) is a positive whole number.
💡 Example 3: find the sum from two terms
The 4th term of an arithmetic sequence is 17 and the 9th term is 42. Find the sum of the first 25 terms.
Step 1: find \(a\) and \(d\) using \(a + (n-1)d\):
\(a + 3d = 17\)
\(a + 8d = 42\)
Subtract: \(5d = 25\), so \(d = 5\), \(a = 2\).
Step 2: apply \(S_n\):
\(S_{25} = \dfrac{25}{2}\bigl[2(2) + (25-1)(5)\bigr]\)
\(= 12.5 \times [4 + 120]\)
\(= 12.5 \times 124 = 1550\)
What's happening?
Two-step problem: solve simultaneous equations to find \(a\) and \(d\), then plug them into the sum formula. Don't try to use the sum formula until you have both values.
💡 Example 4: word problem
Hannah saves £15 in week 1 and increases her saving by £3 each week. How much will she have saved in total after 12 weeks?
This is an arithmetic series with:
\(a = 15\), \(d = 3\), \(n = 12\)
\(S_{12} = \dfrac{12}{2}\bigl[2(15) + (12-1)(3)\bigr]\)
\(= 6 \times [30 + 33]\)
\(= 6 \times 63 = 378\)
Total saved: £378
What's happening?
Translate the words into \(a\), \(d\), \(n\) first. The "total saved" wording is the giveaway that you want \(S_n\) rather than a single term.
🔑 Key points
- The sum formula \(S_n = \tfrac{n}{2}[2a + (n-1)d]\) needs three values: \(a\), \(d\), \(n\).
- Always simplify the contents of the square brackets first, then multiply by \(\tfrac{n}{2}\).
- For "find n given the sum" problems, set \(S_n\) equal to the target and solve the resulting quadratic.
- For "given two terms" problems, find \(a\) and \(d\) first using simultaneous equations.
- For word problems, the phrase "total" or "altogether" usually signals \(S_n\) rather than a specific term.
⚠️ Common pitfalls
- Confusing the sum \(S_n\) with the \(n\)th term itself. The \(n\)th term is one number; \(S_n\) is a total.
- Using \(n\) instead of \(n-1\) inside the bracket. The formula is \((n-1)d\), not \(nd\).
- Forgetting that \(\tfrac{n}{2}\) might be a non-integer (e.g. \(\tfrac{25}{2} = 12.5\)). That is fine; carry on as decimals.
- Solving the quadratic in Example 2 and accepting both roots. Only the positive integer root is valid for \(n\).
- Rounding too early. Keep exact values until the final step, especially with fractional \(\tfrac{n}{2}\).