Arithmetic Sequences (AA SL)
An arithmetic sequence increases (or decreases) by the same fixed amount, the common difference, from one term to the next. This page covers the two formulas you need - for a single term and for the sum of several terms - with worked examples and the mistakes that lose the most marks. It's part of the broader Sequences & Series topic.
23 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference SL1.2. Both formulas come from the formula booklet, so you don't need to memorise them - but you do need to recognise which one a question wants.
nth term
\(u_n = u_1 + (n-1)d\)
Use when you need one specific term - "find the 10th term", "find the first term greater than 100".
Sum of n terms
\(S_n = \dfrac{n}{2}(2u_1 + (n-1)d) = \dfrac{n}{2}(u_1 + u_n)\)
Use when you need a total - "find the sum of the first 20 terms". The second form is faster if you already know \(u_n\).
Need the full syllabus wording and formula-booklet reference table? See Sequences & Series.
Worked examples
An arithmetic sequence has \(u_1 = 5\) and common difference \(d = 3\).
(a) Find \(u_{10}\).
(b) Find \(S_{10}\).
Worked solution
(a) The \(n\)th term of an arithmetic sequence is \(u_n = u_1 + (n-1)d\). M1
\(n=10,\ u_1=5,\ d=3\): \(u_{10} = 5 + (10-1)(3) = 5 + 27 = 32.\) A1
(b) Knowing both \(u_1\) and \(u_{10}\), the efficient choice is \(S_n = \tfrac{n}{2}(u_1+u_n)\). M1
\(S_{10} = \frac{10}{2}(5 + 32) = 5 \times 37 = 185.\) A1
Evaluate \(\displaystyle\sum_{k=1}^{20} (3k - 1)\).
Worked solution
\(3k-1\) is linear in \(k\), so the terms form an AP. At \(k=1\): \(2\); at \(k=20\): \(59\); common difference \(d=3\), \(n=20\). M1
Apply \(S_n=\tfrac{n}{2}(u_1+u_n)\): M1
\(\sum_{k=1}^{20}(3k-1) = \frac{20}{2}(2+59) = 10(61) = 610.\) A1
An arithmetic series has first term \(u_1 = 5\) and last term \(u_n = 95\). The sum of all \(n\) terms is \(750\).
(a)(i) Find \(n\).
(a)(ii) Find the common difference \(d\).
Worked solution
(a)(i) \(S_n = \tfrac{n}{2}(u_1 + u_n) = \tfrac{n}{2}(5 + 95).\) M1
So \(50n = 750\) A1
\(\Rightarrow n = 15.\) A1
(a)(ii) \(u_{15} = u_1 + 14d = 95 \Rightarrow 5 + 14d = 95.\) M1
\(14d = 90\) A1
\(\Rightarrow d = \tfrac{45}{7}.\) A1
Common mistakes
- Using \(n\) instead of \((n-1)\). The nth-term formula multiplies \(d\) by \((n-1)\), not \(n\) - an easy off-by-one error under time pressure.
- Forgetting sigma notation is still an AP. \(\sum_{k=1}^{20}(3k-1)\) is a sum of an arithmetic sequence in disguise - identify \(u_1\), \(d\), and \(n\) first, then use the normal sum formula.
- Picking the wrong sum formula. If you already know the last term, \(S_n=\tfrac{n}{2}(u_1+u_n)\) is faster and less error-prone than expanding \(2u_1+(n-1)d\).
Ready to practise properly?
23 arithmetic-sequence questions, marked instantly like the real exam.
Quick answers
What is the formula for the nth term of an arithmetic sequence?
\(u_n = u_1 + (n-1)d\), where \(u_1\) is the first term and \(d\) is the common difference.
What is the formula for the sum of an arithmetic series?
\(S_n = \tfrac{n}{2}(2u_1 + (n-1)d)\), or equivalently \(S_n = \tfrac{n}{2}(u_1 + u_n)\) once you know the last term.