The big idea: A sequence is a list of numbers.
A series is what you get when you add those numbers together.
So a geometric series is just the sum of the terms of a geometric sequence.
Notation
We write Sₙ for the sum of the first n terms.
- the sum of the first n terms
- the first term
- the nth term (the last one in the sum)
- S₁ is just u₁
- S₂ is u₁ + u₂
- S₃ is u₁ + u₂ + u₃
- …and so on.
Worked example — find S₄ directly
Take the sequence 2, 6, 18, 54, … Find S₄.
Step by step
- List the first 4 terms.
- Add them up.
Final answer
S₄ = 80
Why we need a formula: Adding the terms one by one is fine for small n.
But what if the IB asks for S₅₀?
You'd have to find 50 terms and then add them — way too slow.
That's why the next section gives you a one-step formula.
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The big idea: Instead of adding terms one by one, plug u₁, r, and n into a single formula and you get Sₙ in one step.
There are two equivalent forms — both give the same answer.
- sum of the first n terms
- the first term
- the common ratio
- how many terms you're adding up
Both forms are the same: Multiply the top and bottom of one by −1 and you get the other.
They give the exact same answer.
The only reason to have both is to keep your arithmetic clean — more on that in Section 3.
Only works when r ≠ 1: If r = 1, every term is the same number, so Sₙ is just n × u₁.
The formula has (r − 1) in the denominator, which would be 0 — undefined.
Worked example 1 — r > 1
Find S₆ for the sequence 3, 6, 12, 24, …
Step by step
- Find u₁ and r.
- Since r > 1, use Sₙ = u₁(rⁿ − 1) ÷ (r − 1).
- Work out the bracket.
- Multiply.
- Sanity check by adding.
Final answer
S₆ = 189
Worked example 2 — 0 < r < 1
Find S₅ for the sequence 80, 40, 20, …
Step by step
- Find u₁ and r.
- Since 0 < r < 1, use Sₙ = u₁(1 − rⁿ) ÷ (1 − r).
- Work out the bracket.
- Calculate.
- Sanity check.
Final answer
S₅ = 155
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The big idea: Both forms always give the same answer.
But pick the right one for your value of r and the top and bottom of the fraction will both be positive — which means fewer sign errors and tidier working.
The rule of thumb
| Value of r | Cleaner form | Why |
|---|---|---|
| r > 1 | Sₙ = u₁(rⁿ − 1) ÷ (r − 1) | rⁿ > 1 so (rⁿ − 1) is positive; r > 1 so (r − 1) is positive |
| 0 < r < 1 | Sₙ = u₁(1 − rⁿ) ÷ (1 − r) | rⁿ < 1 so (1 − rⁿ) is positive; r < 1 so (1 − r) is positive |
IB accepts either form.
This rule is just about keeping your working clean.
Same question, both forms — same answer
Sum the first 5 terms of 80, 40, 20, … (so u₁ = 80, r = 0.5).
Step by step
- Using the 'wrong' form (r > 1 form): both top and bottom are negative, so they cancel.
- Using the 'right' form (0 < r < 1 form): top and bottom are both positive — much cleaner.
Final answer
Same answer either way — the second version just has fewer negatives to track.
Watch out — don't drop the −1: A very common slip is to write Sₙ = u₁ × rⁿ ÷ (r − 1).
The numerator must be (rⁿ − 1) or (1 − rⁿ) — not just rⁿ.
For r = 2 and n = 6, rⁿ = 64 but (rⁿ − 1) = 63.
That's a difference of 3 in your final answer (× u₁ = 3 → 3 marks gone).
Don't grab the arithmetic formula by mistake: The arithmetic series formula is Sₙ = (n ÷ 2)(2u₁ + (n − 1)d).
That's for sequences where you add a common difference d.
Geometric series multiply by a common ratio r — they need the geometric sum formula.
If the question says "common ratio", it's geometric.
The big idea: Sometimes IB tells you what Sₙ equals and asks you to find a missing piece — usually n (how many terms) or u₁ (the first term).
The technique is always the same: plug in everything you know, then solve for the missing letter.
Type 1 — find n when Sₙ is known
A geometric sequence has u₁ = 3 and r = 2.
The sum of the first n terms is 189.
Find n.
Step by step
- List what you know.
- Plug into the formula.
- Simplify the denominator.
- Divide both sides by 3.
- Add 1 to both sides.
- Recognise 64 = 2⁶.
Final answer
n = 6
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Type 2 — find u₁ when r, n, and Sₙ are known
A geometric sequence has r = 3 and S₄ = 80.
Find u₁.
Step by step
- Plug what you know into the formula.
- Work out the bracket.
- Divide both sides by 40.
- Check by listing the sequence and adding.
Final answer
u₁ = 2
GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
GDC tip — use the equation solver: When the powers aren't nice (e.g. 2.7ⁿ), your GDC equation solver does the work.
TI-84: MATH → Solver.
Casio fx-CG50: EQUA → SOLVE.
Type the equation, set the unknown to n, and it solves it.
The IB syllabus doesn't expect you to do logs by hand — use the solver.
Examiner trap — exponent is n, not n − 1: In the sum formula the exponent is n (e.g. 2ⁿ).
In the nth-term formula the exponent is n − 1 (e.g. 2ⁿ⁻¹).
Don't mix them up.
If you ever solve for n − 1 in your working (because of how you set things up), add 1 at the end to get the actual answer.