Sequences in Context (AI HL)
Many exam questions don't just hand you \(u_1\) and \(d\) - they describe a situation (a salary, a savings plan, a population) and expect you to build the sequence yourself, or they give you the sum formula \(S_n\) directly and ask you to work backwards to find individual terms. This page covers both skills, with worked examples and the mistakes that lose the most marks. It's part of the broader Sequences & Series topic.
16 questions on this sub-topic.
Two ways to build a term
Covered under IB syllabus reference SL1.2: arithmetic sequences and series, including applications such as simple interest over a number of years, and interpreting a real-life model that isn't perfectly arithmetic.
From a description
\(u_n = u_1 + (n-1)d\)
Read the context to identify the starting value \(u_1\) and the fixed change per step \(d\), then apply the ordinary nth-term formula. Common for salaries, rent, and simple-interest problems.
From a sum formula
\(u_n = S_n - S_{n-1}\ (n>1), \quad u_1 = S_1\)
If you're only given a formula for the running total \(S_n\), the individual term is the difference between consecutive totals. This works even if the resulting sequence turns out not to be arithmetic.
Need the full syllabus wording and formula-booklet reference table? See Sequences & Series.
Worked examples
A worker starts on $32,000 per year, with a raise of $1500 each year.
(a) Find the salary in year 10.
(b) Find the total earned over the first 10 years.
Worked solution
(a) \(u_{10} = 32000 + 9(1500)\) M1
\(= 45500.\) A1
(b) \(S_{10} = \tfrac{10}{2}(32000 + 45500)\) M1
\(= $387\,500.\) A1
The sum of the first \(n\) terms of a sequence is \(S_n = 3n^2 - n\).
(a) Find \(S_3\).
(b) Find an expression for \(u_n\).
Worked solution
(a) Find \(S_3\). \(S_3 = 3(3)^2 - 3 = 27 - 3 = 24.\) A1
(b) (the \(n\)th term is the change in the running total). M1
\(u_n = 3n^2-n - \big(3(n-1)^2-(n-1)\big).\) A1
\(3(n-1)^2-(n-1) = 3n^2-7n+4\), so \(u_n = (3n^2-n)-(3n^2-7n+4) = 6n-4.\) A1
Common mistakes
- Forgetting \(u_1 = S_1\), not \(S_1 - S_0\). The formula \(u_n = S_n - S_{n-1}\) breaks down at \(n=1\) because there's no \(S_0\) term - the first term must be read straight from \(S_1\) instead.
- Assuming a context is arithmetic without checking. Real-life data (bus fares, tiered pricing, rounded percentages) can look arithmetic over a few terms but drift away from a constant difference - state the assumption clearly rather than treating it as fact.
- Skipping the units or context sentence. "In context" answers are often expected in words as well as numbers - stating just "45500" when the question expects "the salary in year 10 is $45,500" can lose an accuracy mark for interpretation.
Ready to practise properly?
16 sequences-in-context questions, marked instantly like the real exam.
Quick answers
How do I recover the nth term from a sum formula S_n?
Use \(u_n = S_n - S_{n-1}\) for \(n>1\), and \(u_1 = S_1\) for the first term. This works whether or not the resulting sequence is arithmetic.
What does it mean for a real-life model to not be perfectly arithmetic?
The values increase by roughly, but not exactly, a constant amount each step - for example rounding, seasonal variation, or a cap. You still use the arithmetic formulas as an approximation, but should comment on where the model breaks down.