IB Maths AI SL Topic 1 — Sequences & Series Paper 1 & 2 In formula booklet ~7 min read

Geometric Sequences & Series

A geometric sequence multiplies by the same number each step. That fixed multiplier is the common ratio, r. Same idea as arithmetic — but with × instead of +.

📘 What you need to know

Finding the nth term

nth term — formula booklet un = u1 × rn − 1
Geometric sequence: u1 = 3, r = 2 3 u1 6 u2 12 u3 24 u4 48 u5 ×2 ×2 ×2 ×2 USE un = u1 × rn − 1 — e.g. find u8 u8 = 3 × 27 = 3 × 128 = 384
Each step multiplies by r. To jump straight to u8 without listing all the doublings, raise r to the (n − 1)th power.

Finding the sum Sn

Sum of n terms — both in formula booklet Sn = u1(rn − 1)r − 1    or    Sn = u1(1 − rn)1 − r
Which version to use? Pick the one that keeps the numerator and denominator positive. For r > 1 the LEFT version (rn − 1) is cleaner. For 0 < r < 1 the RIGHT version (1 − rn) is cleaner. They give the same answer either way.

🧭 Recipe — solve any geometric problem

  1. Identify u1 and r: divide any term by the previous one to get r.
  2. To find a specific term: plug into un = u1 × rn − 1.
  3. Given two terms (not consecutive): divide them — u5/u2 = r3 — then take the appropriate root.
  4. To find a sum: plug into Sn; pick the version that suits your r.
  5. To find n: rearrange to rn = value, then use logs.

Worked examples

WE 1

Find a specific term

A geometric sequence has first term 3 and common ratio 2. Find u8.

Apply u_n = u_1 × r^(n−1) u_8 = 3 × 2^(8−1) = 3 × 2^7 = 3 × 128 u8 = 384
WE 2

Find r and u1 from two non-consecutive terms

In a geometric sequence, u2 = 6 and u5 = 162. Find r and u1.

Divide u_5 by u_2 to eliminate u_1 u_5 / u_2 = (u_1 r⁴) / (u_1 r) = r³ r³ = 162 / 6 = 27 r = ∛27 = 3 Find u_1 from u_2 u_2 = u_1 × r → 6 = u_1 × 3 r = 3, u1 = 2 always DIVIDE consecutive (or wider) terms to find r — never subtract.
WE 3

Decreasing sequence — 0 < r < 1

A geometric sequence has u1 = 100 and r = 0.5. Find u6.

Apply u_n = u_1 × r^(n−1) u_6 = 100 × 0.5^(6−1) = 100 × 0.5^5 = 100 × 0.03125 u6 = 3.125 terms shrink because 0 < r < 1 — sequence halves each time.
WE 4

Sum when r > 1

Find the sum of the first 8 terms of a geometric sequence with u1 = 4 and r = 3.

r > 1, use S_n = u_1(r^n − 1)/(r − 1) S_8 = 4 × (3^8 − 1) / (3 − 1) = 4 × (6561 − 1) / 2 = 4 × 6560 / 2 = 4 × 3280 S8 = 13 120
WE 5

Sum when 0 < r < 1

Find the sum of the first 10 terms of a geometric sequence with u1 = 80 and r = 0.75. Give your answer to 3 s.f.

r < 1, use S_n = u_1(1 − r^n)/(1 − r) S_10 = 80 × (1 − 0.75^10) / (1 − 0.75) = 80 × (1 − 0.0563…) / 0.25 = 80 × 0.9437… / 0.25 = 301.97… S10 ≈ 302 (3 s.f.) use the (1 − r^n) version when r is a fraction or decimal — keeps numbers positive.
WE 6

Real-world — depreciation

A car costs $30 000 new and loses 20% of its value each year. Find its value after 5 years.

Identify geometric sequence loses 20% = keeps 80% of value each year u_1 = $30 000 (start), r = 0.8 After 5 years means 5 multiplications value = 30 000 × 0.8^5 = 30 000 × 0.32768 value = $9830.40 “loses 20%” → multiply by 0.8, NOT by 0.2. The 0.2 is what’s GONE; the 0.8 is what’s LEFT.

💡 Top tips

⚠ Common mistakes

Up next: Applications of Sequences & Series — putting arithmetic and geometric to work on real exam scenarios: compound interest, savings plans, depreciation, populations, and patterns.

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