The big idea: Whenever something changes by the same fixed amount each step — a salary rising by a set raise, seats increasing per row — it is an arithmetic sequence: the start is u₁ and the fixed change is d.
Reading a word problem: "A theatre has 20 seats in the front row and 4 more in each row behind." → u₁ = 20 (front row), d = 4 (extra seats per row).
The 10th row has u₁₀ = 20 + 9(4) = 56 seats.
Translate the words
- "starts at / first / initial" → u₁.
- "each / per / every time" → the common difference d.
- "increases" → d positive; "decreases / falls" → d negative.
- "total / altogether" → a sum Sₙ; "the nth one" → a term uₙ.
Which paper?: These context questions are common on Paper 2 (calculator allowed).
Set up u₁ and d, then let the GDC do the arithmetic.
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When is the sum biggest?: In a decreasing arithmetic sequence the terms eventually turn negative. The sum Sₙ is greatest just before that — at the last term that is still positive or zero.
Find where uₙ = 0; that position gives the maximum sum.
IB-style question — maximum sum
An arithmetic sequence has first term 48 and common difference −3.
Find the maximum value of Sₙ.
Step by step
- Find where the terms reach zero — this is the last term to add.
- So u₁₇ = 0; adding any later (negative) term would only shrink the total. Use the first-and-last form.
- Work it out.
Final answer
The maximum value of Sₙ is S₁₇ = 408.
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
On Paper 2: You can also graph Sₙ or scan a table of Sₙ to read the maximum directly — the IB accepts a graph or table.
Either way, the maximum is at the term where uₙ = 0.
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How many until…?: To find the first term past a limit, set up an inequality with the nth-term formula and solve for n.
Round to the next whole number — n must be a positive integer.
IB-style question — first term below a value
An arithmetic sequence has first term 90 and common difference −7.
Find the first term that is less than 20.
Step by step
- Set the nth term below 20.
- Solve the inequality.
- n must be a whole number, so the first is n = 12. Check.
Final answer
The 12th term (u₁₂ = 13) is the first below 20.
IB-style question — total over a period
A worker is paid 1800 in month 1, rising by 50 each month.
Find the total pay over the first 2 years.
Step by step
- Two years = 24 months, so n = 24 with u₁ = 1800, d = 50.
- Substitute.
- Work it out.
Final answer
Total pay = 57 000.
Common mistakes
- Rounding the wrong way on a threshold inequality.
- Giving n as a decimal — it must be a whole number.
- Using uₙ when the question asks for a total (Sₙ).
Do this instead
- Check which side of the limit the rounded n lands on.
- Round to the next integer and verify the term.
- "Total" means a sum — use Sₙ.