Key Idea: Arithmetic sequences add the same number every step — you'll find a single term or sum a list of terms. It's one of the most reliable Unit-1 questions, on both papers.
🔢 The two formulas
- the first term
- the common difference (the fixed step)
- the position — you add d only (n−1) times
- the sum of the first n terms
- use when you know u₁ and d
- use when you know the first and last term
| You're asked for… | Use | Key move |
|---|---|---|
| d from two terms | gap ÷ steps | u₇ and u₃ are 4 steps apart (7−3), not 7. |
| a term uₙ | uₙ = u₁ + (n−1)d | Spot u₁ first; add d only (n−1) times. |
| a sum Sₙ | Sₙ = (n/2)(…) | The factor is n/2, not n. |
| a term from a sum rule | uₙ = Sₙ − Sₙ₋₁ | Also u₁ = S₁ (Paper 1 favourite). |
| a sigma sum Σ | Sₙ (it's arithmetic) | Terms = upper − lower + 1. |
If the exam gives an explicit rule like uₙ = 20 − 4n, then d = the coefficient of n (= −4) and u₁ = put n = 1 (= 16) — no working needed. A summand linear in the index (e.g. 3r + 2) is always arithmetic, with d = the coefficient of the index.
✏️ IB-style worked examples
IB-style question — find a term from two known terms
The 3rd term of an arithmetic sequence is 17 and the 7th term is 41. Find the 20th term.
Step by step:
Find d: subtract the values, divide by the number of steps (7 − 3).
Step back to u₁: from u₃, subtract two d's.
Now use the nth-term formula with n = 20.
u₂₀ = 119.
IB-style question — recover a term from a sum formula (Paper 1)
The sum of the first n terms of an arithmetic sequence is Sₙ = 2n² + 3n. Find u₁ and an expression for uₙ.
Step by step:
The first term is just S₁.
For any term use uₙ = Sₙ − Sₙ₋₁. Put (n − 1) in place of every n.
Subtract — the 2n² terms cancel, leaving a linear term.
u₁ = 5 and uₙ = 4n + 1 (check: u₁ = 5 ✓).
IB-style question — evaluate a sigma sum (Paper 1)
Without a calculator, evaluate Σr=1¹⁰ (3r + 2).
Step by step:
It's linear in r → arithmetic. First term (put r = 1):
Last term (put r = 10):
Count the terms: upper − lower + 1.
Use the first-and-last sum form.
The sum is 185.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
IB-style question — maximum sum of a decreasing sequence (Paper 2)
An arithmetic sequence has first term 48 and common difference −3. Find the maximum value of Sₙ.
Step by step:
The sum is biggest at the last non-negative term — find where uₙ = 0.
So u₁₇ = 0; adding later (negative) terms only shrinks the total. Use the first-and-last form.
Maximum is S₁₇ = 408.
Important: You add d (n − 1) times, not n times: the 10th term is only 9 steps from u₁. Likewise, two terms are (later − earlier) steps apart, and a sigma sum has (upper − lower + 1) terms — the +1 is the one everyone drops.
Tap each card to reveal the answer.
u₁ = 5, d = 3 — find the 10th term u₁₀ = 32 — add d nine times: 5 + 9×3.
uₙ = 20 − 4n — write down u₁ and d u₁ = 16, d = −4 — put n = 1 for u₁; d is the coefficient of n.
How many terms in Σ from r = 3 to 12? 10 terms — 12 − 3 + 1 (don't forget the +1).
Sum of 4 + 7 + 10 + … (first 10 terms) 175 — (10/2)(2×4 + 9×3) = 5×35.
Sₙ = 2n² + 3n — find u₁ u₁ = S₁ = 5 — substitute n = 1 into Sₙ.
Decreasing sequence — where is the sum biggest? At the last non-negative term — find where uₙ = 0.
Exam Tips
- Two formulas only: uₙ = u₁ + (n−1)d and Sₙ = (n/2)(…). Both are in the booklet.
- Find d from two terms by gap ÷ steps — count steps, not term numbers.
- Explicit rule → d is the coefficient of n; u₁ = put n = 1.
- Sum given as a quadratic? Use u₁ = S₁ and uₙ = Sₙ − Sₙ₋₁.
- Paper 2: sum(seq(…)) totals any Σ, and the table/graph finds a maximum sum.