What is a sequence in the context of IB MYP Mathematics?

In IB MYP Mathematics, a sequence is an ordered list of numbers that follow a specific pattern or rule. Each number in the sequence is called a term. Sequences are a fundamental concept in algebra, helping students understand patterns and relationships between numbers.

How do you find the nth term of an arithmetic sequence?

To find the nth term of an arithmetic sequence, use the formula: nth term = a + (n-1)d, where 'a' is the first term, 'd' is the common difference between consecutive terms, and 'n' is the term number. This formula helps students predict any term in the sequence without listing all previous terms.

What distinguishes a geometric sequence from an arithmetic sequence?

A geometric sequence is characterized by a constant ratio between consecutive terms, known as the common ratio, while an arithmetic sequence has a constant difference between terms. Understanding these differences is crucial for solving problems related to sequences in the IB MYP curriculum.

How can sequences be applied in real-life situations?

Sequences can model various real-life situations, such as calculating interest in finance, predicting population growth, or analyzing patterns in nature. Recognizing these applications helps students appreciate the practical relevance of sequences in mathematics and beyond.

What are some common challenges students face when learning about sequences?

Students often struggle with identifying the pattern or rule governing a sequence, especially when dealing with complex or non-linear sequences. Additionally, applying formulas to find specific terms can be challenging without a solid understanding of the sequence's structure. Practice and familiarity with different types of sequences can help overcome these challenges.

Why is "For quadratic sequences, taking leading coeff = 2nd difference (not half)." a common mistake in Sequences?

Why it happens: Skipping a factor. How to avoid it: Leading coefficient = 12 × constant 2nd difference. Comes from the second derivative of n^2 being 2.

Why is "Using $S_n = \dfrac{a(r^n - 1)}{r - 1}$ when $r = 1$." a common mistake in Sequences?

Why it happens: Applying the formula blindly. How to avoid it: When r = 1, every term equals a. So S_n = na directly. The GP formula is undefined at r = 1.

IB MYP Mathematics (MYP - M)

Sequences

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Short Notes - Sequences

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Detailed Study Notes

Detailed notes on Algebra for IB MYP Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Sequences (Algebra) — IB MYP Mathematics (Extended): finding general formulas algebraically

Now we treat sequences with algebraic rigour — finding $n$ th-term formulas using AP/GP formulae and second-differences for quadratic sequences. This is the core criterion-B skill.

At a glance

Arithmetic: $u_n = a + (n-1)d$ .
Geometric: $u_n = a r^{n-1}$ .
Quadratic sequence: $u_n = an^2 + bn + c$ — constant SECOND differences.
Sum of AP: $S_n = \dfrac{n}{2}(2a + (n-1)d) = \dfrac{n}{2}(a + l)$ .
Sum of GP: $S_n = \dfrac{a(r^n - 1)}{r - 1}$ for $r \ne 1$ .

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Find $n$ th term and sum formulas.
MYP Mathematics A — Identify and find a formula for a quadratic sequence.
MYP Mathematics B — Investigate a sequence to conjecture a rule.
MYP Mathematics C — Justify or prove a general formula.

Finding the $n$ th-term formula

Spot the pattern; write the formula.

Arithmetic case. Spot the COMMON DIFFERENCE $d$ .

$u_n = a + (n-1)d.$

Worked example. Sequence $4, 11, 18, 25, \ldots$ . Find $u_n$ .

$a = 4$ , $d = 7$ .
$u_n = 4 + (n-1) \times 7 = 4 + 7n - 7 = 7n - 3$ .
Check: $u_1 = 4$ . ✓ $u_2 = 11$ . ✓

Geometric case. Spot the COMMON RATIO $r$ .

$u_n = a r^{n-1}.$

Worked example. Sequence $5, 15, 45, 135, \ldots$ . Find $u_n$ .

$a = 5$ , $r = 3$ .
$u_n = 5 \times 3^{n-1}$ .

Quadratic case (Extended). If SECOND differences are constant.

Worked example. Sequence $3, 7, 13, 21, 31, \ldots$ .

First differences: $4, 6, 8, 10$ (not constant — not AP).
Second differences: $2, 2, 2$ (constant — QUADRATIC).
Formula: $u_n = an^2 + bn + c$ with $a = \tfrac{\text{2nd diff}}{2} = 1$ .
So $u_n = n^2 + bn + c$ . Use two terms to find $b, c$ .
$n = 1$ : $1 + b + c = 3 \Rightarrow b + c = 2$ .
$n = 2$ : $4 + 2b + c = 7 \Rightarrow 2b + c = 3$ .
Subtract: $b = 1$ , then $c = 1$ .
Formula: $u_n = n^2 + n + 1$ . Check: $u_3 = 9 + 3 + 1 = 13$ . ✓

AP: $u_n = a + (n-1)d$ .
GP: $u_n = a r^{n-1}$ .
Quadratic: constant SECOND differences. Leading coefficient = $\tfrac{\text{2nd diff}}{2}$ .

Sums of arithmetic and geometric sequences

Two beautiful closed-form formulas.

Sum of an arithmetic sequence (AP):

$S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}(a + l)$

where $l = u_n$ is the LAST term.

Why? Write the sum forward and backwards, add them: every pair sums to $(a + l)$ , and there are $n$ pairs.

Worked example. Find $S_{20}$ for $4, 11, 18, 25, \ldots$

$a = 4$ , $d = 7$ , $n = 20$ .
$S_{20} = \dfrac{20}{2}(2 \times 4 + 19 \times 7) = 10 \times (8 + 133) = 1410$ .

Sum of a geometric sequence (GP):

$S_n = \frac{a(r^n - 1)}{r - 1}, \quad r \ne 1.$

Worked example. Find $S_6$ for $2, 6, 18, 54, \ldots$ (GP with $a = 2$ , $r = 3$ ).

$S_6 = \dfrac{2(3^6 - 1)}{3 - 1} = \dfrac{2(728)}{2} = 728$ .

For $|r| < 1$ , an INFINITE GP converges to $S_\infty = \dfrac{a}{1 - r}$ (this is touched on at Extended level, fully developed in DP Maths).

$S_n^{AP} = \dfrac{n}{2}(a + l)$ or $\dfrac{n}{2}(2a + (n-1)d)$ .
$S_n^{GP} = \dfrac{a(r^n - 1)}{r - 1}$ .
Infinite GP: $S_\infty = \dfrac{a}{1-r}$ if $|r| < 1$ .

Investigating patterns (criterion B)

MYP's signature skill: spot → conjecture → test → prove.

MYP criterion B specifically tests sequence/pattern investigation. The expected process:

Compute several terms of the pattern.
Spot the pattern: AP? GP? Quadratic? Other?
Conjecture a formula.
Test the formula on new values.
Justify or prove the formula (the high-mark step).

Worked example. Investigate the pattern: pile of cans where each row has one more can than the row above. 1 can on top, 2 next, 3 next, ... Find a formula for the total number of cans in an $n$ -row pile.

First few values: 1, 3, 6, 10, 15. (Triangle numbers!)
First differences: 2, 3, 4, 5 (not constant).
Second differences: 1, 1, 1 (constant — quadratic).
Leading coeff: $\tfrac{1}{2}$ . So $u_n = \tfrac{1}{2}n^2 + bn + c$ .
$n = 1$ : $\tfrac{1}{2} + b + c = 1 \Rightarrow b + c = \tfrac{1}{2}$ .
$n = 2$ : $2 + 2b + c = 3 \Rightarrow 2b + c = 1$ .
Subtract: $b = \tfrac{1}{2}$ , so $c = 0$ .
Formula: $u_n = \tfrac{1}{2}n^2 + \tfrac{1}{2}n = \dfrac{n(n+1)}{2}$ .

Justification: The total number of cans is $1 + 2 + 3 + \ldots + n$ , which is the standard sum of the first $n$ integers. Using AP sum: $a = 1$ , $d = 1$ , $l = n$ : $S = \tfrac{n}{2}(1 + n) = \dfrac{n(n+1)}{2}$ . ✓

A criterion-B answer should INCLUDE this last 'why does the formula make sense' step — not just present the formula.

Spot → conjecture → test → prove.
For quadratic: use 2nd differences.
Justification matters as much as the formula.

Quick recap

AP: $u_n = a + (n-1)d$ . Sum: $S_n = \tfrac{n}{2}(a+l)$ .
GP: $u_n = a r^{n-1}$ . Sum: $S_n = \dfrac{a(r^n - 1)}{r - 1}$ .
Quadratic: constant 2nd differences, leading coeff = $\tfrac{1}{2}$ × 2nd diff.
Criterion B: spot → conjecture → test → justify.

Memorise this

Verbatim phrases and definitions MYP criterion-A markschemes credit.

AP sum: $S = \tfrac{n}{2}(a + l)$
GP sum: $S = \dfrac{a(r^n - 1)}{r - 1}$
Sum of first $n$ integers: $\tfrac{n(n+1)}{2}$
Quadratic: 2nd differences

How it’s examined

Criterion A: apply formulas. Criterion B: investigate a pattern. Criterion C: clearly justify the formula. Criterion D: model a real-world sequence.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

Take this whole topic with you

Step-by-step worked examples — Sequences

Step-by-step solutions to past-paper-style questions on sequences, written exactly the way a tutor would explain them at the board.

1Find an AP formula
Building confidence• AP
Question
Find a formula for the $n$ th term of $9, 4, -1, -6, \ldots$ .
Step-by-step solution
1. Step 1
  $a = 9$ , $d = -5$ .
2. Step 2
  $u_n = 9 + (n-1)(-5) = 9 - 5n + 5 = 14 - 5n$ .
3. Step 3
  Check: $u_1 = 14 - 5 = 9$ . ✓ $u_4 = 14 - 20 = -6$ . ✓
Answer
$u_n = 14 - 5n$
2Sum of an AP
Building confidence• AP sum
Question
Find the sum of the first 30 terms of $5, 8, 11, 14, \ldots$ .
Step-by-step solution
1. Step 1
  AP with $a = 5$ , $d = 3$ , $n = 30$ .
2. Step 2
  $S_n = \dfrac{n}{2}(2a + (n-1)d) = \dfrac{30}{2}(10 + 29 \times 3)$ .
3. Step 3
  $= 15 \times (10 + 87) = 15 \times 97 = 1455$ .
Answer
$S_{30} = 1455$
3Find a quadratic formula
Stretch• quadratic sequence
Question
Find a formula for the sequence $2, 7, 14, 23, 34, \ldots$
Step-by-step solution
1. Step 1
  1st differences: $5, 7, 9, 11$ . 2nd differences: $2, 2, 2$ (constant — quadratic).
2. Step 2
  Leading coefficient: $\tfrac{2}{2} = 1$ . So $u_n = n^2 + bn + c$ .
3. Step 3
  $n=1$ : $1 + b + c = 2 \Rightarrow b + c = 1$ .
4. Step 4
  $n=2$ : $4 + 2b + c = 7 \Rightarrow 2b + c = 3$ .
5. Step 5
  Subtract: $b = 2$ , so $c = -1$ .
6. Step 6
  $u_n = n^2 + 2n - 1$ . Check: $u_5 = 25 + 10 - 1 = 34$ . ✓
Answer
$u_n = n^2 + 2n - 1$
4Sum of a GP
Stretch• GP sum
Question
Find the sum of the first 8 terms of $3, 6, 12, 24, \ldots$ .
Step-by-step solution
1. Step 1
  GP with $a = 3$ , $r = 2$ , $n = 8$ .
2. Step 2
  $S_n = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{3(2^8 - 1)}{2 - 1} = 3(256 - 1) = 3 \times 255 = 765$ .
Answer
$S_8 = 765$

Key Definitions and Keywords — Sequences

Definitions to memorise and the exact keywords mark schemes credit for sequences answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Sum of arithmetic series $S_n$
Examiner keyword
$S_n = \dfrac{n}{2}(2a + (n-1)d) = \dfrac{n}{2}(a + l)$ .
Sum of geometric series $S_n$
Examiner keyword
$S_n = \dfrac{a(r^n - 1)}{r - 1}$ for $r \ne 1$ .
Quadratic sequence
A sequence whose $n$ th term is of the form $an^2 + bn + c$ . Has constant SECOND differences.
Recursive definition
A definition that gives each term in terms of previous ones, e.g. $u_n = u_{n-1} + u_{n-2}$ for Fibonacci.

Common Mistakes and Misconceptions — Sequences

The traps other students keep falling into on sequences questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕For quadratic sequences, taking leading coeff = 2nd difference (not half).
Why it happens
Skipping a factor.
How to avoid it
Leading coefficient = $\tfrac{1}{2}$ × constant 2nd difference. Comes from the second derivative of $n^2$ being 2.
✕Using $S_n = \dfrac{a(r^n - 1)}{r - 1}$ when $r = 1$ .
Why it happens
Applying the formula blindly.
How to avoid it
When $r = 1$ , every term equals $a$ . So $S_n = na$ directly. The GP formula is undefined at $r = 1$ .

Sequences — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Sequences

Short Notes - Sequences

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Find an AP formula

2Sum of an AP

3Find a quadratic formula

4Sum of a GP

Sum of arithmetic series SnS_nSn​

Sum of geometric series SnS_nSn​

Quadratic sequence

Recursive definition

✕For quadratic sequences, taking leading coeff = 2nd difference (not half).

✕Using Sn=a(rn−1)r−1S_n = \dfrac{a(r^n - 1)}{r - 1}Sn​=r−1a(rn−1)​ when r=1r = 1r=1.

Sequences — frequently asked questions

Sum of arithmetic series $S_n$

Sum of geometric series $S_n$

✕Using $S_n = \dfrac{a(r^n - 1)}{r - 1}$ when $r = 1$ .