IB Maths
Paper 1 & 2
18 min read
Arithmetic Sequences & Series
An arithmetic sequence adds (or subtracts) the same number each time. Just 2 formulas handle nearly every IB question on this topic. Let’s go.
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What you need to know
- How to spot an arithmetic sequence (look for a constant difference)
- The nth term formula to find any term
- The sum formula to add the first n terms
- How to handle “find u1 and d” using simultaneous equations
What is an arithmetic sequence?
A sequence where each term increases (or decreases) by the same amount — called the common difference, d.
Quick examples:
- 3, 7, 11, 15, 19, … d = +4
- 20, 17, 14, 11, 8, … d = −3 (decreasing)
- 2, 2.5, 3, 3.5, 4, … d = +0.5
- −5, −2, 1, 4, 7, … d = +3
How to find d: subtract any term from the one after it.
d = u2 − u1 = u3 − u2 = u4 − u3 = …
Formula 1: The nth term
un = u1 + (n − 1)d
Where:
- u1 = first term
- d = common difference
- n = position of the term you want
This formula is in the IB formula booklet. No need to memorise — just know how to use it.
Quick example: find u20 for the sequence 3, 7, 11, 15, …
- u1 = 3, d = 4
- u20 = 3 + (20 − 1)(4)
- u20 = 3 + 76 = 79
Formula 2: The sum of the first n terms
You get two versions in the formula booklet — pick whichever fits the question.
Sn = n2 [2u1 + (n − 1)d]
— use when you know u1 and d
Sn = n2 (u1 + un)
— use when you know the first and last terms
Quick example: sum of 3 + 7 + 11 + 15 + … up to 20 terms
- u1 = 3, d = 4, n = 20
- S20 = 202 [2(3) + (20 − 1)(4)]
- S20 = 10 [6 + 76] = 10 × 82 = 820
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Tips
- Both sum formulas are in the formula booklet
- If a question gives you the last term, use Sn = n2(u1 + un) — it’s faster
- Hard questions often give you two pieces of info (e.g. u4 and u9) — use simultaneous equations
Worked Examples
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Example 1 — Find a specific term
Find u15 of the sequence 5, 9, 13, 17, …
Answer:
u1 = 5, d = 4
u15 = 5 + (15 − 1)(4)
= 5 + 56
u15 = 61
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Example 2 — Find u1 and d using simultaneous equations
The 4th term of an arithmetic sequence is 10 and the 9th term is 25. Find u1 and d.
Answer:
Step 1: write each term using the formula un = u1 + (n − 1)d.
u4 = u1 + 3d = 10
u9 = u1 + 8d = 25
Step 2: subtract to eliminate u1.
5d = 15 → d = 3
Step 3: substitute back.
u1 + 3(3) = 10 → u1 = 1
u1 = 1, d = 3
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Example 3 — Find d when Sn is given
The sum of the first 10 terms of an arithmetic sequence is 630, and the first term is 18. Find d.
Answer:
Use Sn = (n/2)[2u1 + (n − 1)d] with n = 10, u1 = 18, S10 = 630.
630 = (10/2)[2(18) + 9d]
630 = 5(36 + 9d)
126 = 36 + 9d
9d = 90
d = 10
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Example 4 — Find u1 when d and Sn are given
Same sum (S10 = 630), but now d = 11. Find the first term.
Answer:
630 = (10/2)[2u1 + 9(11)]
630 = 5(2u1 + 99)
126 = 2u1 + 99
2u1 = 27
u1 = 13.5
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Example 5 — Find how many terms (n)
An arithmetic sequence has first term 7 and common difference 5. How many terms are needed for the sum to exceed 500?
Answer:
Set up the inequality Sn > 500.
(n/2)[2(7) + (n − 1)(5)] > 500
(n/2)[14 + 5n − 5] > 500
(n/2)(5n + 9) > 500
5n² + 9n − 1000 > 0
Solve using GDC: n > 13.4… (reject negative root)
n must be a whole number, so n = 14.
n = 14 terms
Check: S13 = 481, S14 = 546 ✓
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Common mistakes
- Using n instead of (n − 1) in the formula. The first term has n = 1, but (n − 1)d = 0 then — that’s correct!
- Wrong sign for d. Decreasing sequences have negative d. For 20, 17, 14, … d = −3 (not 3).
- Forgetting to halve. The sum formula has n2 at the front. Don’t drop it.
- Mixing up the formulas. Use Sn = n2(u1 + un) only when you know the LAST term, not just d.
- Counting terms wrong. From u5 to u15 is 11 terms, not 10. (15 − 5 + 1 = 11)
Final word: just 2 formulas, both in the booklet. Practise spotting which one to use — that’s where the real exam skill lies. Hard questions usually need simultaneous equations or solving a quadratic in n.
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