Key Idea: A geometric sequence multiplies by the same ratio r each step — the engine behind compound interest, populations and bouncing balls. It runs through both papers: by-hand on Paper 1, GDC table / finance solver on Paper 2.
🔢 The two formulas
- first term
- common ratio — what you × by each step
- term number (you ×r only n−1 times)
- sum of the first n terms
- use (rⁿ − 1)/(r − 1)
- use (1 − rⁿ)/(1 − r)
| You're given… | Do this |
|---|---|
| u₁ and r | Substitute straight into uₙ or Sₙ. |
| Two terms (e.g. u₂, u₅) | r = (steps)-th root of value-ratio. Even gap ⇒ r = ±. |
| Three terms with an unknown | Set middle² = product (u₂² = u₁u₃) and solve. |
| A % change in context | r = 1 + rate (growth) or 1 − rate (decay); value = A₀rⁿ. |
Geometric multiplies (× r); arithmetic adds (+ d). Divide consecutive terms — same ratio ⇒ geometric. 3, 6, 12, 24 (× 2) is geometric; 3, 6, 9, 12 (+ 3) is arithmetic.
✏️ IB-style worked examples
IB-style question — find the nth term
A geometric sequence has u₃ = 18 and u₆ = 486. Find an expression for the nth term uₙ.
Step by step:
Find r: divide the later value by the earlier, then take the (steps)-th root. The gap 6 − 3 = 3 steps.
Step back to the first term: u₃ = u₁r², so divide by r².
Put u₁ = 2 and r = 3 into the formula.
uₙ = 2 × 3ⁿ⁻¹.
IB-style question — show a closed form for the sum (Paper 1)
A geometric sequence has u₁ = 4 and r = 3. Show that the sum of the first n terms is Sₙ = 2(3ⁿ − 1).
Step by step:
Write the sum formula.
Substitute u₁ = 4 and r = 3.
Cancel to reach the required form.
Sₙ = 2(3ⁿ − 1), as required.
IB-style question — depreciation by a fixed % (decay)
A machine worth $20 000 loses 15% of its value each year. After how many whole years is it first worth less than $8000?
Step by step:
Decay, so the ratio is 1 − the rate.
Value after n years = start × rⁿ. Set it below 8000.
Solve (logs or GDC) and round n up.
After 6 years.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: You ×r one fewer time than the term number. 2016 → 2025 lists 10 years but only 9 increases happen; the 5th bounce is the 6th term (5 jumps, the drop is term 1). Sketch one arrow per step and count to the term you want.
Tap each card to reveal the answer.
u₁ = 2, r = 3 — find the 5th term uₙ = u₁rⁿ⁻¹ = 2 × 3⁴ = 162 (multiply by r only 4 times).
u₁ = 3, u₃ = 12 — find r Value-ratio 4, gap 2 steps (even) → r = ±√4 = ±2; take +2 if terms are positive.
x, x + 6, x + 18 are geometric — find x Middle² = product: (x+6)² = x(x+18) → x = 6.
u₁ = 5, r = 3 — sum of the first 6 terms S₆ = 5(3⁶ − 1)/(3 − 1) = 5(728)/2 = 1820.
Salary $52 000, rising 4%/yr — translate to r Growth → r = 1 + 0.04 = 1.04; value = 52000 × 1.04ⁿ.
A ball is dropped from 12 m and rebounds to ½ each time. How far has it travelled by the time it lands for the 4th time? Count up to the 4th landing: the drop (12 m), then 3 rebounds, each travelled up and back down. The rebound heights 6, 3, 1.5 add to 10.5 (a 3-term sum), so total = 12 + 2 × 10.5 = 33 m. (It stops at the 4th landing — a finite sum, no infinity needed; the forever version is 1.8.)
Exam tips
- uₙ = u₁rⁿ⁻¹ — multiply by r only (n − 1) times; the first term is already there.
- Two terms → r = the (steps)-th root of the value-ratio; an even gap means r could be ±.
- Three terms geometric ⇒ middle² = product; if that gives a quadratic, report both values.
- Sₙ needs u₁, r and n — pick the form that keeps top and bottom positive; "show" questions: simplify until it matches.
- Context: r = 1 ± rate, value = A₀rⁿ; for "how long until" round n up and use the GDC table or TVM solver.