The big idea: Money, populations and values that change by the same percentage each period form a geometric sequence.
The common ratio is r = 1 + rate for growth, or r = 1 − rate for decay.
For example, 6% growth → r = 1.06; 15% decay → r = 0.85.
Translate the words
- "grows / increases by x%" → r = 1 + x/100.
- "falls / depreciates by x%" → r = 1 − x/100.
- "doubles / triples" → solve rⁿ = 2 or 3.
- "value after t periods" → start × rᵗ.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Compound interest is geometric: A classic Paper 2 question asks how long money takes to reach a target — model the balance as start × rⁿ and solve for n (logs or the TVM solver).
IB-style question — when does it double?
$2000 is invested at 6% per year, compounded annually.
After how many whole years does it first exceed $4000?
Step by step
- As a geometric sequence, the balance after n years is start × rⁿ.
- Divide by 2000 — it doubles.
- Solve (logs or the GDC) and round up.
Final answer
12 years.
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Two ways, same answer: On Paper 2 you can use the TVM solver (above) or the geometric way (solve rⁿ = 2 with logs).
Both give n ≈ 11.9 → 12 years. Always round up for "how long until".
Practice with real exam questions
Answer exam-style questions and get AI feedback that shows you exactly what examiners want to see in a full-marks response.
Decay multiplies by less than 1: When something shrinks by the same percentage each period (depreciation, cooling), the ratio is r = 1 − rate (between 0 and 1) — then model and solve exactly like growth.
IB-style question — depreciation
A machine worth $20 000 loses 15% of its value each year.
(a) Find its value after 4 years. (b) After how many whole years is it first worth less than $8000?
Step by step
- Decay ratio.
- (a) Value after 4 years = start × r⁴.
- (b) Set below 8000 and solve, then round up.
Final answer
(a) ≈ $10 440. (b) After 6 years.
Rounding for decay: Still round n up for "how many years until below $X".
Because the value keeps a fixed percentage each year, it shrinks but never quite reaches zero.