IB Math AA SL Topic 1.2: Arithmetic sequences & series | Notes & Flashcards | Aimnova

IB Math AA SL — Arithmetic sequences & series

Topic 1.2 of IB Mathematics: Analysis and Approaches covers Arithmetic sequences & series, which is part of Unit 1: Number & Algebra. Students explore key concepts including nth term, Sum of n terms, Sigma notation, Applications. A strong understanding of arithmetic sequences & series is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Arithmetic sequences & series

Key Idea: Arithmetic sequences add the same number every step — you'll find a single term or sum a list of terms. It's one of the most reliable Unit-1 questions, on both papers.

🔢 The two formulas

u_n = u_1 + (n-1)d

$u_1$: the first term
$d$: the common difference (the fixed step)
$n$: the position — you add d only (n−1) times

S_n = \frac{n}{2}\big(2u_1 + (n-1)d\big) = \frac{n}{2}\big(u_1 + u_n\big)

$S_n$: the sum of the first n terms
$2u_1+(n-1)d$: use when you know u₁ and d
$u_1+u_n$: use when you know the first and last term

If the exam gives an explicit rule like uₙ = 20 − 4n, then d = the coefficient of n (= −4) and u₁ = put n = 1 (= 16) — no working needed. A summand linear in the index (e.g. 3r + 2) is always arithmetic, with d = the coefficient of the index.

✏️ IB-style worked examples

IB-style question — find a term from two known terms

The 3rd term of an arithmetic sequence is 17 and the 7th term is 41. Find the 20th term.

Step by step:

Find d: subtract the values, divide by the number of steps (7 − 3).
$d = \frac{41 - 17}{7 - 3} = \frac{24}{4} = 6$
Step back to u₁: from u₃, subtract two d's.
$u_1 = 17 - 2(6) = 5$
Now use the nth-term formula with n = 20.
$u_{20} = 5 + (20-1)\times 6 = 5 + 114$

Final answer:

u₂₀ = 119.

IB-style question — recover a term from a sum formula (Paper 1)

The sum of the first n terms of an arithmetic sequence is Sₙ = 2n² + 3n. Find u₁ and an expression for uₙ.

Step by step:

The first term is just S₁.
$u_1 = S_1 = 2(1)^2 + 3(1) = 5$
For any term use uₙ = Sₙ − Sₙ₋₁. Put (n − 1) in place of every n.
$S_{n-1} = 2(n-1)^2 + 3(n-1) = 2n^2 - n - 1$
Subtract — the 2n² terms cancel, leaving a linear term.
$u_n = (2n^2 + 3n) - (2n^2 - n - 1) = 4n + 1$

Final answer:

u₁ = 5 and uₙ = 4n + 1 (check: u₁ = 5 ✓).

IB-style question — evaluate a sigma sum (Paper 1)

Without a calculator, evaluate Σr=1¹⁰ (3r + 2).

Step by step:

It's linear in r → arithmetic. First term (put r = 1):
$u_1 = 3(1)+2 = 5$
Last term (put r = 10):
$u_{10} = 3(10)+2 = 32$
Count the terms: upper − lower + 1.
$n = 10 - 1 + 1 = 10$
Use the first-and-last sum form.
$S_{10} = \frac{10}{2}(5 + 32) = 5 \times 37$

Final answer:

The sum is 185.

IB-style question — maximum sum of a decreasing sequence (Paper 2)

An arithmetic sequence has first term 48 and common difference −3. Find the maximum value of Sₙ.

Step by step:

The sum is biggest at the last non-negative term — find where uₙ = 0.
$48 - 3(n-1) = 0 \;\Rightarrow\; n = 17$
So u₁₇ = 0; adding later (negative) terms only shrinks the total. Use the first-and-last form.
$S_{17} = \frac{17}{2}(48 + 0)$

Final answer:

Maximum is S₁₇ = 408.

Important: You add d (n − 1) times, not n times: the 10th term is only 9 steps from u₁. Likewise, two terms are (later − earlier) steps apart, and a sigma sum has (upper − lower + 1) terms — the +1 is the one everyone drops.

Tap each card to reveal the answer.

Exam Tips

Two formulas only: uₙ = u₁ + (n−1)d and Sₙ = (n/2)(…). Both are in the booklet.
Find d from two terms by gap ÷ steps — count steps, not term numbers.
Explicit rule → d is the coefficient of n; u₁ = put n = 1.
Sum given as a quadratic? Use u₁ = S₁ and uₙ = Sₙ − Sₙ₋₁.
Paper 2: sum(seq(…)) totals any Σ, and the table/graph finds a maximum sum.

IB Math AA SL — Arithmetic sequences & series

Key concepts in Arithmetic sequences & series

Key Idea: Arithmetic sequences add the same number every step — you'll find a single term or sum a list of terms. It's one of the most reliable Unit-1 questions, on both papers.

🔢 The two formulas

u_n = u_1 + (n-1)d

$u_1$: the first term
$d$: the common difference (the fixed step)
$n$: the position — you add d only (n−1) times

S_n = \frac{n}{2}\big(2u_1 + (n-1)d\big) = \frac{n}{2}\big(u_1 + u_n\big)

$S_n$: the sum of the first n terms
$2u_1+(n-1)d$: use when you know u₁ and d
$u_1+u_n$: use when you know the first and last term

If the exam gives an explicit rule like uₙ = 20 − 4n, then d = the coefficient of n (= −4) and u₁ = put n = 1 (= 16) — no working needed. A summand linear in the index (e.g. 3r + 2) is always arithmetic, with d = the coefficient of the index.

✏️ IB-style worked examples

IB-style question — find a term from two known terms

The 3rd term of an arithmetic sequence is 17 and the 7th term is 41. Find the 20th term.

Step by step:

Find d: subtract the values, divide by the number of steps (7 − 3).
$d = \frac{41 - 17}{7 - 3} = \frac{24}{4} = 6$
Step back to u₁: from u₃, subtract two d's.
$u_1 = 17 - 2(6) = 5$
Now use the nth-term formula with n = 20.
$u_{20} = 5 + (20-1)\times 6 = 5 + 114$

Final answer:

u₂₀ = 119.

IB-style question — recover a term from a sum formula (Paper 1)

The sum of the first n terms of an arithmetic sequence is Sₙ = 2n² + 3n. Find u₁ and an expression for uₙ.

Step by step:

The first term is just S₁.
$u_1 = S_1 = 2(1)^2 + 3(1) = 5$
For any term use uₙ = Sₙ − Sₙ₋₁. Put (n − 1) in place of every n.
$S_{n-1} = 2(n-1)^2 + 3(n-1) = 2n^2 - n - 1$
Subtract — the 2n² terms cancel, leaving a linear term.
$u_n = (2n^2 + 3n) - (2n^2 - n - 1) = 4n + 1$

Final answer:

u₁ = 5 and uₙ = 4n + 1 (check: u₁ = 5 ✓).

IB-style question — evaluate a sigma sum (Paper 1)

Without a calculator, evaluate Σr=1¹⁰ (3r + 2).

Step by step:

It's linear in r → arithmetic. First term (put r = 1):
$u_1 = 3(1)+2 = 5$
Last term (put r = 10):
$u_{10} = 3(10)+2 = 32$
Count the terms: upper − lower + 1.
$n = 10 - 1 + 1 = 10$
Use the first-and-last sum form.
$S_{10} = \frac{10}{2}(5 + 32) = 5 \times 37$

Final answer:

The sum is 185.

IB-style question — maximum sum of a decreasing sequence (Paper 2)

An arithmetic sequence has first term 48 and common difference −3. Find the maximum value of Sₙ.

Step by step:

The sum is biggest at the last non-negative term — find where uₙ = 0.
$48 - 3(n-1) = 0 \;\Rightarrow\; n = 17$
So u₁₇ = 0; adding later (negative) terms only shrinks the total. Use the first-and-last form.
$S_{17} = \frac{17}{2}(48 + 0)$

Final answer:

Maximum is S₁₇ = 408.

Important: You add d (n − 1) times, not n times: the 10th term is only 9 steps from u₁. Likewise, two terms are (later − earlier) steps apart, and a sigma sum has (upper − lower + 1) terms — the +1 is the one everyone drops.

Tap each card to reveal the answer.

Exam Tips

Two formulas only: uₙ = u₁ + (n−1)d and Sₙ = (n/2)(…). Both are in the booklet.
Find d from two terms by gap ÷ steps — count steps, not term numbers.
Explicit rule → d is the coefficient of n; u₁ = put n = 1.
Sum given as a quadratic? Use u₁ = S₁ and uₙ = Sₙ − Sₙ₋₁.
Paper 2: sum(seq(…)) totals any Σ, and the table/graph finds a maximum sum.

IB Math AA SL — Arithmetic sequences & series

Key concepts in Arithmetic sequences & series

🔢 The two formulas

✏️ IB-style worked examples

IB-style question — find a term from two known terms

IB-style question — recover a term from a sum formula (Paper 1)

IB-style question — evaluate a sigma sum (Paper 1)

IB-style question — maximum sum of a decreasing sequence (Paper 2)

Exam Tips

What you'll learn in Topic 1.2

Study resources — 1.2 Arithmetic sequences & series

nth term

Sum of n terms

Sigma notation

Applications

Ready to study Arithmetic sequences & series?

Ready to practice?

IB Math AA SL — Arithmetic sequences & series

Key concepts in Arithmetic sequences & series

🔢 The two formulas

✏️ IB-style worked examples

IB-style question — find a term from two known terms

IB-style question — recover a term from a sum formula (Paper 1)

IB-style question — evaluate a sigma sum (Paper 1)

IB-style question — maximum sum of a decreasing sequence (Paper 2)

Exam Tips

What you'll learn in Topic 1.2

Study resources — 1.2 Arithmetic sequences & series

nth term

Sum of n terms

Sigma notation

Applications

Ready to study Arithmetic sequences & series?

Ready to practice?