IB Maths AA SL Topic 1 — Number & Algebra Paper 1 & 2 šŸŽÆ Skill ~4 min practice

AA SL Geometric Series Sums skills

A geometric series multiplies each term by the same ratio. There are two formulas — one for finite sums (Sn) and one for infinite sums (Sāˆž). The trick is choosing correctly and checking convergence before you reach for the infinite version.

The Method

GP: u1, u1r, u1r², u1r³, … u1 = first term Ā· r = common ratio Ā· n = number of terms
Formula 1

Finite Sum Sn

Sn = u1(rn āˆ’ 1)r āˆ’ 1
use for the first n terms
Formula 2

Infinite Sum Sāˆž

Sāˆž = u11 āˆ’ r
use when sum is infinite Ā· only if |r| < 1
āš ļø

Infinite sum exists ONLY when |r| < 1

If |r| ≄ 1 the terms don’t shrink and the series doesn’t converge. Always state |r| < 1 before you apply Sāˆž = u1 / (1 āˆ’ r).

Three steps every time

  1. Identify what you have. Pick out u1, r, and either n (finite) or “āˆž” (infinite). r = any term Ć· previous term.
  2. Choose the formula. Finite “first n terms”? → Formula 1. “Sum to infinity”? → Check |r| < 1 first, then Formula 2.
  3. Plug in carefully. Watch the order of (rn āˆ’ 1) and (r āˆ’ 1) — both numerator and denominator have r minus 1, not 1 minus r.

Worked examples

WE 1 EASY

Find the sum of the first 8 terms of: 3, 6, 12, 24, …

step 1 — identify u₁ = 3, r = 6 Ć· 3 = 2, n = 8 step 2 — Formula 1 (finite) Sā‚ˆ = 3(2⁸ āˆ’ 1) / (2 āˆ’ 1) = 3(256 āˆ’ 1) / 1 = 3 Ɨ 255 Sā‚ˆ = 765 r = (any term) Ć· (previous term) — pick any pair to find it!
WE 2 MEDIUM

Find the sum to infinity of: 12, 4, 4/3, 4/9, …

step 1 — identify u₁ = 12, r = 4 Ć· 12 = ā…“ step 2 — check convergence |ā…“| < 1 āœ“ → Sāˆž exists step 3 — Formula 2 Sāˆž = 12 / (1 āˆ’ ā…“) = 12 / (ā…”) = 12 Ɨ (3/2) = 18 Sāˆž = 18 always state |r| < 1 before applying Sāˆž — it’s a marking criterion!
WE 3 HARD

A geometric series has u1 = 5 and r = 2. Find the smallest value of n for which Sn > 1000.

step 1 — set up the inequality 5(2ⁿ āˆ’ 1) / (2 āˆ’ 1) > 1000 5(2ⁿ āˆ’ 1) > 1000 2ⁿ āˆ’ 1 > 200 2ⁿ > 201 step 2 — solve using logs n log 2 > log 201 n > log 201 / log 2 ā‰ˆ 7.65 step 3 — round UP to next integer n must be a whole number ≄ 7.65 smallest n = 8 “smallest n such that Sn > k” — always round UP, never down!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Identify r first — that tells you which formula to use.
Q1 EASY Find the sum of the first 6 terms of: 2, 6, 18, 54, … Show answer ā–¼Hide answer ā–²
u₁ = 2, r = 3, n = 6 → Formula 1 S₆ = 2(3⁶ āˆ’ 1) / (3 āˆ’ 1) = 2(729 āˆ’ 1) / 2 = 728 S₆ = 728
Q2 EASY Find the sum to infinity of: 16, 8, 4, 2, … Show answer ā–¼Hide answer ā–²
u₁ = 16, r = ½ — |r| < 1 āœ“ Sāˆž = 16 / (1 āˆ’ ½) = 16 / (½) Sāˆž = 32
Q3 MEDIUM A GP has u₁ = 80 and r = āˆ’Ā½. Find Sāˆž. Show answer ā–¼Hide answer ā–²
|āˆ’Ā½| = ½ < 1 āœ“ → converges Sāˆž = 80 / (1 āˆ’ (āˆ’Ā½)) = 80 / (3/2) = 80 Ɨ (2/3) Sāˆž = 160/3 negative r is fine — it’s |r| that matters for convergence!
Q4 MEDIUM Find the sum of the first 10 terms of: 100, 50, 25, 12.5, … Show answer ā–¼Hide answer ā–²
u₁ = 100, r = ½, n = 10 → Formula 1 S₁₀ = 100((½)¹⁰ āˆ’ 1) / (½ āˆ’ 1) = 100(1/1024 āˆ’ 1) / (āˆ’Ā½) = 100(āˆ’1023/1024) / (āˆ’Ā½) = 200 Ɨ 1023/1024 = 199.8 (4 sf) S₁₀ ā‰ˆ 199.8 use the GDC for messy fractions — no shame in it on Paper 2!
Q5 HARD A GP has u₁ = 4 and Sāˆž = 10. Find r. Show answer ā–¼Hide answer ā–²
step 1 — set up 10 = 4 / (1 āˆ’ r) step 2 — solve for r 10(1 āˆ’ r) = 4 10 āˆ’ 10r = 4 10r = 6 r = 3/5 = 0.6 check: |0.6| < 1 āœ“ → series does converge as required

⚠ Common mistakes

šŸ“–

Want the theory?

Read the full Geometric Sequences & Series notes for the link to un = u1rnāˆ’1, why convergence requires |r| < 1, and applications to compound interest and depreciation.

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