The big idea: To add the first n terms of a geometric sequence, use the sum formula — it needs u₁, r and n.
Which form?: Use (rⁿ − 1)/(r − 1) when r > 1, and (1 − rⁿ)/(1 − r) when 0 < r < 1.
Either gives the same answer — it just keeps top and bottom positive.
Plug in u₁, r and n: A common exam question gives you the sequence and asks for a sum — put u₁, r and n into the formula. If r is not given, find it first.
IB-style question — sum with r given
A geometric sequence has u₁ = 5 and r = 3.
Find the sum of the first 6 terms.
Step by step
- Write the sum formula (r > 1, so use the first form).
- Substitute u₁ = 5, r = 3, n = 6.
- Simplify.
Final answer
S₆ = 1820.
IB-style question — find r first
A geometric sequence has first term 4 and second term 12.
Find the sum of the first 8 terms.
Step by step
- Find r — divide consecutive terms.
- Write the formula, then substitute u₁ = 4, r = 3, n = 8.
- Simplify.
Final answer
S₈ = 13120.
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Tidy the sum into the target form: A Paper 1 favourite asks you to show the sum equals a tidy closed form like a(bⁿ − 1) — substitute into the formula and simplify until it matches.
IB-style question — show a closed form
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.
Final answer
Sₙ = 2(3ⁿ − 1), as required.
IB-style question — when r is a fraction
A geometric sequence has u₁ = 9 and r = ⅓.
Show that Sₙ = (27/2)(1 − 3⁻ⁿ).
Step by step
- r < 1, so use the second form.
- Substitute u₁ = 9 and r = ⅓.
- Dividing by ⅔ means multiplying by 3/2.
Final answer
Sₙ = (27/2)(1 − 3⁻ⁿ), as required.
Match the printed form: A "show" question gives you the answer — your marks are for the steps. Keep simplifying until your expression looks exactly like the printed form (same powers, same fractions).
Smallest n past a target: A Paper 2 favourite asks for the smallest n whose sum passes a target — set Sₙ above the target and read the GDC table, rounding up.
IB-style question — smallest n (Paper 2)
A geometric sequence has first term 80 and common ratio 1.5.
Find the smallest value of n for which the sum of the first n terms exceeds 20 000.
Step by step
- Write the sum formula and tidy it — this is what goes into the calculator.
- Enter it in the GDC as Y₁ (use X for n), so the table shows the running total after each term.
- Read down the table to the first row past 20 000 — it crosses between n = 11 and n = 12.
Final answer
Smallest n = 12 (and S₁₁ ≈ 13 680 is still under — the check the examiner wants).
GDC tip (Paper 2): Put the sum formula in Y₁, then open the table (2nd → GRAPH) and scroll until Y₁ first passes the target.
Take the first n that is over — and check the row just before is still under. No logs needed.
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Down once, then up-and-down each bounce: A dropped ball travels the drop once, then each rebound twice (up, then down). So total distance = drop + 2 × (sum of the rebound heights), and the rebounds are a geometric sum.
IB-style question — distance to the 4th bounce
A ball is dropped from 12 m and rebounds to ½ of its height each time.
Find the total distance travelled by the time it hits the ground for the 4th time.
Step by step
- Set it up: the drop once, plus twice the rebound sum. The rebounds are geometric, so this is what you substitute into.
- Read off what goes in: the first rebound is half of 12, so u₁ = 6 and r = ½. The 4th time it lands is after 3 rebounds, so n = 3.
- Add the three rebound heights (6, 3, 1.5) with the sum formula.
- Drop counted once, rebounds counted twice.
Final answer
33 m.
Don't forget to double: The classic slip is counting each rebound once. Every rebound is climbed and fallen, so the rebound sum is doubled — only the first drop is counted once.