HL AA · Functions

Sequences, series, and induction proofs (IB Math AA HL)

Updated October 2026

Quick answer: Arithmetic: uₙ = u₁ + (n − 1)d and Sₙ = (n/2)(2u₁ + (n − 1)d). Geometric: uₙ = u₁rⁿ⁻¹ and Sₙ = u₁(1 − rⁿ)/(1 − r), with sum to infinity u₁/(1 − r) when |r| < 1. AA HL combines these with sigma notation, inequalities in n, and proof by induction of sum formulas.

HL AA combines sequences with proof by induction — 'prove that the sum of the first n squares is n(n+1)(2n+1)/6'.

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Key ideas

Worked example

A geometric series has first term 12 and sum to infinity 48. Find the common ratio, and the least number of terms for which the sum is greater than 47.

  1. S∞ = 12/(1 − r) = 48, so 1 − r = 1/4 and r = 3/4 (and |r| < 1, as needed).
  2. Sₙ = 12(1 − 0.75ⁿ)/(1 − 0.75) = 48(1 − 0.75ⁿ).
  3. 48(1 − 0.75ⁿ) > 47 gives 0.75ⁿ < 1/48.
  4. n > ln(1/48) ÷ ln 0.75 ≈ 13.5, so n = 14.
  5. Check: S₁₃ ≈ 46.86 and S₁₄ ≈ 47.14.

Answer: r = 3/4; 14 terms.

Exam-style HL AA questions

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  1. Question 1 of 5
  2. Question 2 of 5
  3. Question 3 of 5
  4. Question 4 of 5
  5. Question 5 of 5

5 quick questions on arithmetic sequences. No sign-in needed.

An arithmetic sequence has first term \(5\) and common difference \(3\). Find the \(20\)th term.
Choose your answer · keys 1–4

Do it on your GDC

A calculator is allowed on AA Paper 2 and Paper 3, not on Paper 1. Keystrokes for the TI-84 Plus CE, TI-Nspire CX II and Casio fx-CG50, with a checked worked example:

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FAQ

When does a geometric series have a sum to infinity?

Only when the common ratio satisfies |r| < 1, so the terms shrink towards 0. The sum is then u₁/(1 − r). If |r| ≥ 1 the partial sums grow without limit or keep oscillating.

Can I use my calculator for sequences, series, and induction proofs?

In HL AA, Paper 1 is non-calculator and the other papers allow a GDC, so learn the exact method as well as the calculator one. The Sequences and sums guide gives the keystrokes for the TI-84 Plus CE, TI-Nspire CX II and Casio fx-CG50.

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