The big idea: If something changes by the same percentage every period (every year, every hour, every bounce), it's a geometric model. The keyword is constant percentage change — not a constant amount.
Real-world examples you'll see in IB
| Situation | What changes each step | Growth or decay? |
|---|---|---|
| Bacteria doubling every hour | × 2 each hour | Growth (r = 2) |
| Population growing 4% per year | × 1.04 each year | Growth (r = 1.04) |
| Car loses 15% value per year | × 0.85 each year | Decay (r = 0.85) |
| Radioactive substance halves every 10 years | × 0.5 every 10 years | Decay (r = 0.5) |
Quick check — is it geometric?: Ask yourself: "Is each step a percentage of the previous step?" If yes → geometric. If each step adds the same amount (e.g. "saves $50 each month") → arithmetic, not geometric.
Geometric vs arithmetic — the real-world test
| Phrase in the question | Type |
|---|---|
| "Increases by 5% per year" | Geometric (r = 1.05) |
| "Increases by $200 per year" | Arithmetic (d = 200) |
| "Halves every 8 hours" | Geometric (r = 0.5) |
| "Drops by 10 units per second" | Arithmetic (d = −10) |
Watch out — "%" doesn't always mean geometric: If a question says "adds 5% of the original starting value each year" — that's a fixed amount each year (5% of u₁), so it's arithmetic. Geometric requires the percentage to be of the current value, not the original.
The big idea: Once you know it's geometric, the job is always the same — two steps:__LINEBREAK___Step 1. Find r (the multiplier per period) from the percentage in the question.__LINEBREAK___Step 2. Plug u₁ (starting value), r, and n (which period you want) into:__LINEBREAK___uₙ = u₁ · rⁿ⁻¹__LINEBREAK__The two formulas below do all of step 1 for you.
Step 1 — Finding r from the percentage
- the common ratio (the multiplier per period)
- the percentage change per period (just the number — e.g. 4 for 4%)
| Phrase | p | r |
|---|---|---|
| Grows 4% per year | 4 | 1 + 4/100 = 1.04 |
| Grows 25% per month | 25 | 1 + 25/100 = 1.25 |
| Decays 12% per year | 12 | 1 − 12/100 = 0.88 |
| Loses 3% per second | 3 | 1 − 3/100 = 0.97 |
Step 2 — Plug r into the nth-term formula
- the value at period n — what you're finding
- the starting value (given in the question)
- the multiplier per period — the value you just found using r = 1 + p/100 or r = 1 − p/100
- which term you want: n = 1 is the start, n = 2 is after 1 period, n = 7 is after 6 periods
Worked example — population growth
A town has 5000 people and grows 4% per year. How many people will there be after 6 years?
Step by step
- Step 1: find r from the percentage.
- Step 2: identify u₁ and n. "After 6 years" means multiplying by r six times, so the exponent is 6.
- Calculate (use your GDC).
- Round to a sensible number of people.
Final answer
About 6327 people after 6 years.
The off-by-one trap on n: The exponent on r equals the number of periods that have passed — the number of times you've multiplied by r. So "after 6 years" → exponent is 6. In the formula uₙ = u₁ · rⁿ⁻¹, that means n = 7 (since 7 − 1 = 6). Either way: count how many times r was applied, and that's your exponent.
Cleaner version (per-period exponent): A lot of textbooks write the model as value after t periods = u₁ · rᵗ — where t is just "how many periods passed". This is the same formula with the off-by-one already absorbed. Use whichever feels safer, as long as the exponent matches the number of times r has been applied.
Worked example — depreciation
A car worth $24 000 loses 15% of its value every year. What is it worth after 4 years?
Step by step
- Step 1: find r. 15% loss → decay.
- Step 2: after 4 years → exponent is 4.
- Calculate.
Final answer
About $12 528 after 4 years.
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The big idea: IB doesn't just want the number. It wants the number with units, rounded properly, and interpreted in context. "u₆ = 6326.6" is not a finished answer — "about 6327 people" is.
The 3 things to check on every context answer
- Units — people, dollars, grams, bacteria, $/year. Always include them.
- Rounding — IB default is 3 significant figures unless the question says otherwise. Read carefully — "to the nearest dollar" is different from "3 s.f.".
- Sense check — does your answer sound right? A car worth $50 million after 4 years probably means you forgot decay. A bacteria count of 0.3 means you used the wrong r.
| Question phrasing | How to round |
|---|---|
| No instruction given | 3 significant figures (IB default) |
| "Give your answer to the nearest dollar" | Whole dollars |
| "Give your answer correct to 2 decimal places" | Two decimals |
| "How many full bottles can be filled?" | Round down (you can't have 3.7 full bottles) |
| "How many years until the value drops below…" | Round up (you need to pass the threshold) |
Watch out — context dictates rounding direction: If the question asks "after how many years does the population exceed 10 000?" and you get n = 6.4, the answer is n = 7 (you round up — at year 6 it's still under 10 000). For "how many full crates can you fill from 47.8 crates of stock?", the answer is 47 (round down). Don't blindly round to the nearest whole number — think about what the situation requires.
Worked example — full context answer
A bacterial colony starts at 200 cells and grows 25% per hour. After 5 hours, how many bacteria are there?
Step by step
- Set up: u₁ = 200, r = 1.25, multiply by r five times for 5 hours.
- Calculate.
- Round — bacteria are whole cells, so round to a whole number.
Final answer
About 610 bacteria after 5 hours.
Write a sentence: Method marks come from showing your formula and substitution. The final accuracy mark often goes only to a clear final statement: "After 5 hours there are approximately 610 bacteria." Don't leave the answer as "610" with no context — the examiner won't infer it for you.
The big idea: IB application questions almost always come in 3 parts: (a) find r from the situation, (b) find a future value, (c) find when the value reaches some target. Once you know this pattern, you can predict what's coming.
| Part | What it usually asks | What you do |
|---|---|---|
| (a) | Find r from a percentage | Use r = 1 + p/100 (or 1 − p/100) |
| (b) | Find the value after n periods | Use uₙ = u₁ · rⁿ⁻¹ (or u₁ · rᵗ for t periods passed) |
| (c) | Find when the value first reaches/exceeds X | Set up the equation, solve for n with your GDC, round in the right direction |
Full 3-part walkthrough
A new 15 000?
Step by step
- (a) 12% loss → decay.
- (b) After 5 years means we multiply by r five times.
- (b) Calculate.
- (b) State with units, 2dp for money.
- (c) Set up the inequality. We want value < 15000.
- (c) Divide both sides by 35000.
- (c) Use your GDC equation solver, or test values: 0.88⁶ ≈ 0.464 (still above), 0.88⁷ ≈ 0.409 (below).
- (c) State in context.
Final answer
(a) r = 0.88 (b) About $18 470.62 (c) After 7 years.
Watch out — "after how many years" needs rounding up: If your GDC gives n ≈ 6.85 in part (c), the answer is 7, not 6. At year 6 the value is still above $15 000 — you need one more whole year to dip below. Always check whether the question wants "first reaches" or "first drops below" and round accordingly.
GDC shortcut for part (c): For "first reaches/drops below X" questions, the fastest method on most GDCs is: enter the formula in Y₁ = u₁ × rˣ (or your relevant model), then use TABLE to scroll x = 1, 2, 3, … and read off the first row that crosses your threshold. No solver, no logs, no algebra.