IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 ~10 min read

Geometric Sequences & Series

A geometric sequence is one where you multiply by the same number to get from each term to the next. That fixed multiplier is the common ratio r. The structure mirrors arithmetic — there’s an n-th term formula and a sum formula — but with one bonus: when |r| < 1, the terms shrink fast enough that even an infinite sum has a finite value. This sum to infinity is unique to geometric, and it’s a favourite Paper 1 topic.

📘 What you need to know

Geometric = constant ratio. Each term is the previous one times r.
n-th term: u_n = u₁ · r^{n − 1} ✓ in the formula booklet.
Sum of first n terms — two equivalent forms (both in the booklet):
S_n = u₁(rⁿ − 1)r − 1 (use when r > 1)
S_n = u₁(1 − rⁿ)1 − r (use when r < 1)
Sum to infinity: S_∞ = u₁1 − r, valid only when |r| < 1. ✓ in the booklet.
Behaviour depends on r: increases for r > 1, decreases for 0 < r < 1, alternates sign when r < 0.
To find r from two terms: divide one by the other. The u₁ cancels and leaves r raised to a power.

What is a geometric sequence?

The defining property is simple: divide any term by the previous term and you always get the same number. That’s the common ratio r.

Examples: 3, 6, 12, 24, 48, … (r = 2) | 16, 8, 4, 2, 1, … (r = ½) | 5, −10, 20, −40, … (r = −2)

Three behaviour types — based on r

📈 Increasing

r > 1

3, 6, 12, 24, 48, …

terms grow without bound

📉 Decreasing

0 < r < 1

16, 8, 4, 2, 1, …

terms shrink towards 0

🔄 Alternating

r < 0

5, −10, 20, −40, …

flips sign each step

Three different geometric sequences

To check whether a sequence is geometric, divide consecutive terms: u₂/u₁, then u₃/u₂, then u₄/u₃. If they’re all equal, you have a geometric sequence and that ratio is r.

The n-th term formula

To get from u₁ to u_n, multiply by r a total of (n − 1) times.

n-th term of a geometric sequence u_n = u₁ · r^{n − 1} ✓ in formula booklet

🤔 Why n − 1, not n?

For the same reason as the arithmetic version. u₁ hasn’t been multiplied yet — it’s the starting point. u₂ has been multiplied by r once, u₃ twice, and so on. The number of multiplications is one less than the term number.

Quick check: for the sequence 3, 6, 12, 24, … (u₁ = 3, r = 2), the 10th term is u₁₀ = 3 · 2⁹ = 3 · 512 = 1536.

The “divide” trick — finding r from two terms

If you’re given two terms (say u_m and u_k), divide one by the other. The u₁ cancels and you’re left with a single power of r:

u_ku_m = u₁ r^{k − 1}u₁ r^{m − 1} = r^{k − m}

Two formulas for the sum

Both formulas give the same answer, but one is cleaner depending on whether r is bigger or smaller than 1.

📈

Form 1 — for r > 1

S_n = u₁(rⁿ − 1)r − 1

Use when r > 1 — keeps both numerator and denominator positive.

✓ in formula booklet

📉

Form 2 — for r < 1

S_n = u₁(1 − rⁿ)1 − r

Use when r < 1 — same reason, keeps signs tidy.

✓ in formula booklet

Don’t worry about which to “officially” use — both give the same answer for any value of r ≠ 1. Pick whichever leaves you with positive numbers in the numerator and denominator. The formula booklet has both.

The sum to infinity

Here’s where geometric sequences do something arithmetic sequences can’t: when the terms shrink fast enough (when |r| < 1), the sum of all infinitely many terms approaches a finite value. We call this S_∞ (S-infinity), and it’s the limit of S_n as n → ∞.

The convergence condition

Converges (finite sum) |r| < 1 (i.e. −1 < r < 1)

Diverges (no finite sum) |r| ≥ 1 (i.e. r ≥ 1 or r ≤ −1)

The formula

Sum to infinity (when |r| < 1) S_∞ = u₁1 − r ✓ in formula booklet, with condition |r| < 1

🤔 Why does this work?

Look at Form 2 of the partial sum: S_n = u₁(1 − rⁿ)/(1 − r). When |r| < 1, the term rⁿ shrinks to 0 as n grows large. So the formula reduces to u₁(1 − 0)/(1 − r) = u₁/(1 − r). When |r| ≥ 1, rⁿ doesn’t shrink — it grows or oscillates, and the sum has no limit.

🌍 A real-world feel for it

Walk halfway across a room (½ a unit). Then halfway across the remaining gap (¼). Then halfway again (⅛). After infinitely many steps you’ve travelled ½ + ¼ + ⅛ + … = 1 — exactly across the room. The geometric series with u₁ = ½, r = ½ has S_∞ = (½)/(1 − ½) = 1. ✓

Always state the convergence condition before using S_∞. IB papers expect you to write something like “since |r| < 1, the series converges, so S_∞ = …” — that line earns a method mark.

Worked examples

WE 1

Find a specific term

A geometric sequence has first term 4 and common ratio 3. Find the 8th term.

Substitute into u_n = u₁·r^{n − 1} u₈ = 4 · 3⁷ = 4 · 2187 u₈ = 8748

WE 2

Find r and u₁ using the divide trick

The 4th term of a geometric sequence is 24 and the 7th term is 192. Find the first term and the common ratio.

Step 1: Divide u₇ by u₄ — the u₁‘s cancel u₇u₄ = u₁r⁶u₁r³ = r³ = 19224 = 8 Step 2: Solve for r r³ = 8 → r = 2 Step 3: Find u₁ using u₄ u₁ · 2³ = 24 → 8u₁ = 24 → u₁ = 3 u₁ = 3, r = 2 dividing kills u₁ — same idea as subtracting in arithmetic, but with division instead

WE 3

Sum with r > 1 (Form 1)

Find the sum of the first 8 terms of the geometric sequence with u₁ = 5 and r = 2.

Step 1: r > 1 so use Form 1 S₈ = 5(2⁸ − 1)2 − 1 Step 2: Compute = 5(256 − 1)1 = 5 × 255 S₈ = 1275

WE 4

Sum with r < 1 (Form 2)

Find the sum of the first 6 terms of the geometric sequence with u₁ = 64 and r = ½.

Step 1: r < 1 so use Form 2 S₆ = 64(1 − (½)⁶)1 − ½ Step 2: Compute (½)⁶ = 1/64 = 64(1 − 1/64)½ = 64 × (63/64)½ = 63½ = 126 S₆ = 126

WE 5

Sum to infinity

The first three terms of a geometric sequence are 12, 4, 4/3, … Show that the series converges and find S_∞.

Step 1: Find r r = u₂u₁ = 412 = 13 Step 2: Check convergence condition |r| = 1/3 < 1 ✓ → series converges Step 3: Apply the formula S_∞ = u₁1 − r = 121 − 1/3 = 122/3 = 12 × 32 = 18 S_∞ = 18 always state |r| < 1 first — that’s a method mark

WE 6

Find the smallest n such that S_n exceeds a value (using logs)

A geometric sequence has u₁ = 4 and r = 3. Find the smallest value of n for which S_n > 5000.

Step 1: Set up the inequality with Form 1 4(3ⁿ − 1)3 − 1 > 5000 2(3ⁿ − 1) > 5000 3ⁿ − 1 > 2500 3ⁿ > 2501 Step 2: Take logs and solve n · ln(3) > ln(2501) n > ln(2501)/ln(3) ≈ 7.12 Step 3: Smallest integer n satisfying this n = 8 always check: S₇ = 4372 < 5000, S₈ = 13120 > 5000 ✓

💡 Top tips

Spot it first: if consecutive terms have a constant ratio, it’s geometric. If they have a constant difference, it’s arithmetic.
Two terms given → divide. Dividing one by the other eliminates u₁ and leaves r raised to a power.
Pick the sum form to keep signs positive. If r > 1, use Form 1. If r < 1, use Form 2. They give the same answer either way.
For S_∞, always check |r| < 1 first. State it explicitly — IB awards a method mark for the convergence statement.
If finding n, you’ll need logs. Take ln (or log) of both sides, use the power law to bring n down, then solve.
If you find r^{even power} = positive number, remember r can be positive or negative. Both roots may need to be considered.
For non-calculator papers, keep fractions exact. Don’t decimalise (½)⁶ as 0.015625 — leave it as 1/64.

⚠ Common mistakes

Using rⁿ instead of r^{n − 1}. The 5th term involves r⁴, not r⁵.
Applying S_∞ when |r| ≥ 1. The series diverges — there’s no finite limit. The formula doesn’t apply.
Confusing geometric with arithmetic. “Common ratio” = multiplicative; “common difference” = additive. The formulas are different.
Sign errors with negative r. Sequences with r < 0 alternate sign — easy to miscalculate u_n.
Forgetting to take both roots when r², r⁴, etc. give a positive number. Both ± may be valid.
Not stating the convergence condition for S_∞ questions. Even if you have the right answer, you’ll lose the method mark without that line.
Putting r < 1 instead of |r| < 1. The convergence condition is on the magnitude — r = −0.5 also converges, even though it’s “less than 1”.

Geometric sequences show up everywhere — population modelling, compound interest, radioactive decay, infinite repeating decimals. The formulas are quick once you’ve drilled them. The bonus topic (sum to infinity) is one of the most elegant single ideas in pre-university maths — make sure you can explain why it works, not just plug in.

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