IB Maths AI SL Topic 1 — Sequences & Series Paper 1 & 2 Real-world ~7 min read

Applications of Sequences & Series

First decide whether the situation is arithmetic (adds/subtracts the same amount) or geometric (multiplies/divides by the same factor). Then pick the right formula and plug in. That’s the whole skill.

📘 How to spot which is which

Arithmetic — same amount added or subtracted each step. Words like “increases by $X”, “gets Y more each year”, “X fewer each row”.
Geometric — multiplied or divided by the same factor each step. Words like “X% per year”, “doubles”, “halves”, “grows at rate r”.
Percentages → multipliers: +5% means × 1.05; −20% means × 0.8; “loses 8%” means × 0.92.
Half-life = the time for the quantity to halve, so r = 0.5 per half-life period.
“Total accumulated” → use S_n. “How much in year/period n” → use u_n.
“When does it reach/exceed X?” → solve for n. Often needs logs (geometric) or a quadratic (arithmetic sum).

Arithmetic or geometric? — decide first

Identify the type FIRST, then pick the right formula. Mixing them up is the most common cause of lost marks on these questions.

🧭 Recipe — any application problem

Identify the type: fixed amount → arithmetic; fixed factor or percentage → geometric.
Find u₁ and d or r from the question.
Decide what’s asked: a specific term (u_n), a total (S_n), or a time/position (n).
Plug into the right formula; use the GDC for tricky arithmetic.
Round sensibly: money 2 d.p., people / years whole numbers, otherwise 3 s.f.

Worked examples

WE 1

Arithmetic — salary raises

Maya starts a job at $42 000 per year and receives a $2 500 raise each year. (a) Find her salary in year 10. (b) Find her total earnings over the first 10 years.

Identify (arithmetic — fixed raise) u_1 = 42 000, d = 2 500 (a) Year-10 salary — use u_n u_10 = 42 000 + 9(2 500) = 42 000 + 22 500 = $64 500 (b) Total over 10 years — use S_n S_10 = 10/2 × (2(42 000) + 9(2 500)) = 5 × (84 000 + 22 500) = 5 × 106 500 (a) $64 500 (b) $532 500

WE 2

Arithmetic — stacking pattern

A display stacks cans in rows. The bottom row has 18 cans, and each row above has 2 fewer cans. The stack has 9 rows. How many cans are used in total?

Identify (arithmetic, decreasing — fixed loss per row) u_1 = 18, d = −2, n = 9 Use S_n S_9 = 9/2 × (2(18) + 8(−2)) = 9/2 × (36 − 16) = 9/2 × 20 S₉ = 90 cans check the top row: u_9 = 18 + 8(−2) = 2 cans ✓

WE 3

Geometric — compound interest

$8 000 is deposited in a bank account that pays 6% compound interest per year. Find the balance after 12 years, to 2 d.p.

Identify (geometric — fixed % each year) multiplier = 1 + 6% = 1.06 After 12 years (12 multiplications) balance = 8000 × 1.06^12 = 8000 × 2.0122… balance = $16 097.57 +6% means × 1.06, never × 0.06.

WE 4

Geometric decay — radioactive

A 500 g sample of a radioactive substance decays at 8% per year. Find its mass after 15 years, to 3 s.f.

Identify (geometric, decreasing — fixed % loss) loses 8% = keeps 92% each year u_1 = 500, r = 0.92 After 15 years mass = 500 × 0.92^15 = 500 × 0.2863… = 143.15… mass ≈ 143 g (3 s.f.) “loses 8%” → multiplier is 0.92 (what’s left), not 0.08.

WE 5

Geometric — half-life

A drug dose of 240 mg has a half-life of 8 hours in the body. How much remains after 32 hours?

Identify (geometric — halves every period) multiplier per 8 hours: r = 0.5 Count half-lives in 32 hours 32 ÷ 8 = 4 half-lives Apply amount = 240 × 0.5^4 = 240 × 0.0625 amount = 15 mg match the multiplier to the period — r = 0.5 per HALF-LIFE, not per hour.

WE 6

Find when a population exceeds a target (uses logs)

A town has 12 000 people and grows at 3% per year. After how many whole years will the population first exceed 20 000?

Set up (geometric, r = 1.03) 12 000 × 1.03^n > 20 000 1.03^n > 20 000 / 12 000 = 5/3 Solve using logs n > log(5/3) / log(1.03) n > 17.28… Smallest whole n at n = 17: 12 000 × 1.03^17 = 19 834 (still ≤ 20 000) at n = 18: 12 000 × 1.03^18 = 20 429 (exceeds) n = 18 years always check the integer either side — the question asks for whole years.

💡 Top tips

Read for the keyword: “fixed amount” / “the same each” → arithmetic. “Per cent” / “rate” / “factor” / “doubles” / “halves” → geometric.
Simple vs compound interest: simple interest is arithmetic (fixed $ each year); compound is geometric (×(1 + rate)).
Percent change → multiplier: +x% means × (1 + x/100); −x% means × (1 − x/100).
Match index to the question: “after 5 years” usually means 5 multiplications (geometric) or after the 5th term (arithmetic). Sketch the first 2 terms to check.

⚠ Common mistakes

Using arithmetic for a percentage: “grows by 5% each year” is GEOMETRIC, not arithmetic.
Using the percent itself as r: 8% loss → r = 0.92, NOT 0.08.
Off-by-one in the exponent: “after 5 years” with compound interest typically means × r⁵ from year 0 — not r⁴.
Forgetting to round properly: money to 2 d.p., people / years to a whole number, otherwise 3 s.f.

That wraps Sequences & Series. Up next in AI SL Topic 1: Financial Mathematics — compound interest done properly, with all the formula-booklet variations, plus annuities and amortisation.

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