What is a sequence in Mathematics, and how is it relevant to the Cambridge IGCSE curriculum?

In Mathematics, a sequence is an ordered list of numbers that follow a specific pattern or rule. Sequences are a fundamental concept in the Cambridge IGCSE curriculum as they help students understand patterns and relationships in numbers, which are crucial for solving algebraic problems and developing logical reasoning skills.

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

To find the nth term of an arithmetic sequence, you use the formula: nth term = a + (n-1)d, where 'a' is the first term and 'd' is the common difference between consecutive terms. This formula is essential in the Cambridge IGCSE Mathematics curriculum for solving problems related to sequences and series.

What is the difference between arithmetic and geometric sequences in Algebra?

An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant, known as the common difference. In contrast, a geometric sequence is one where each term is found by multiplying the previous term by a fixed, non-zero number called the common ratio. Understanding these differences is crucial for Cambridge IGCSE students as it helps in identifying and solving sequence-related problems.

Can you explain how to derive the formula for the nth term of a geometric sequence?

The nth term of a geometric sequence can be found using the formula: nth term = a * r^(n-1), where 'a' is the first term and 'r' is the common ratio. This formula is derived by recognizing that each term is a multiple of the previous term by the common ratio. Mastery of this concept is important for students studying the Cambridge IGCSE Mathematics syllabus.

Why is understanding sequences and the nth term important in Algebra and real-world applications?

Understanding sequences and the nth term is vital in Algebra because it forms the basis for more complex mathematical concepts such as series, calculus, and financial mathematics. In real-world applications, sequences are used in areas like computer science, economics, and engineering to model and predict patterns and trends. For Cambridge IGCSE students, this knowledge is foundational for both academic and practical problem-solving.

Why is "Writing $T_{n} = dn$ instead of $T_{n} = dn + (a - d)$" a common mistake in Sequences and nth Term?

Why it happens: Students stop at the common difference and forget the constant correction. How to avoid it: After identifying d, find the constant: T_1 = a tells you what T_n = dn + c should give for n = 1.

Why is "Using the second difference as the coefficient of $n^{2}$ (instead of half)" a common mistake in Sequences and nth Term?

Why it happens: Easy to forget the divide-by-two step. How to avoid it: Coefficient of n^2 = HALF the second difference. Always. Source: 0580/42 — examiner reports

Why is "Starting the index at $n = 0$" a common mistake in Sequences and nth Term?

Why it happens: Programming habits leak in. How to avoid it: In Cambridge IGCSE, n starts at 1. The first term is T_1, not T_0.

Why is "Not verifying the rule against the given terms" a common mistake in Sequences and nth Term?

Why it happens: Time pressure. How to avoid it: Substitute n = 1, 2, 3 into your rule and confirm the values match the given sequence.

Cambridge IGCSE Mathematics (0580)

Sequences and nth Term

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Short Notes - Sequences and nth Term

Detailed Study Notes

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

Sequences and the nth Term — Cambridge IGCSE 0580 Maths Extended (2026)

Linear, quadratic, geometric and special sequences. Find the nth term, predict any term, identify the type, and continue patterns. The nth term is the formula that compresses a sequence into a single rule.

At a glance

Linear (arithmetic): constant 1st difference $d$ . Formula: $a_{n} = a_{1} + (n - 1)d$ .
Quadratic: constant 2nd difference. Formula: $a_{n} = An^{2} + Bn + C$ .
Geometric: constant ratio $r$ . Formula: $a_{n} = a_{1} \cdot r^{n-1}$ .
Special sequences: square numbers $n^{2}$ , cube numbers $n^{3}$ , triangular numbers $\dfrac{n(n+1)}{2}$ .
Find term-from-formula: substitute $n$ into the formula.
Find term number from value: set formula = value and solve for $n$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.13 — Continue a given number sequence; recognise patterns; find the formula for the nth term.
Recognise quadratic, cubic and geometric sequences.

Linear (arithmetic) sequences

Constant first difference. Formula: $a_{n} = a_{1} + (n - 1) d$ , or by inspection.

A sequence is linear (arithmetic) when the gap between consecutive terms is constant. That gap is the common difference $d$ .

Worked. $5, 8, 11, 14, 17, \ldots$ .

First term $a_{1} = 5$ . Common difference $d = 3$ .
$a_{n} = 5 + (n - 1)(3) = 3n + 2$ .
Check: $n = 1 \Rightarrow 5$ ✓. $n = 2 \Rightarrow 8$ ✓. $n = 4 \Rightarrow 14$ ✓.

Quick formula. For any linear sequence: $a_{n} = dn + (a_{1} - d)$ . The coefficient of $n$ is $d$ ; the constant is the "zeroth term" you'd get if you went one step BEFORE $a_{1}$ .

Worked. Find the nth term of $7, 4, 1, -2, \ldots$ .

$d = -3$ . $a_{1} = 7$ . Zeroth term: $7 - (-3) = 10$ .
$a_{n} = -3n + 10$ .

Use. "What's the 50th term?" → $a_{50} = -3(50) + 10 = -140$ . "Which term equals $-50$ ?" → $-3n + 10 = -50 \Rightarrow n = 20$ .

A constant first difference d means the nth term is aₙ = dn + c.

Constant 1st difference = linear sequence.
$a_{n} = dn + c$ where $d$ is the difference and $c$ is the zeroth term.
Substitute $n$ to find any term.
Solve $a_{n} =$ value to find term number.

Quadratic sequences

Constant SECOND difference. The leading coefficient is half that 2nd difference.

A sequence is quadratic when the SECOND difference is constant.

Method to find the nth term.

Compute the 1st and 2nd differences.
Leading coefficient $A = \dfrac{1}{2}(\text{2nd difference})$ .
Subtract $An^{2}$ from each term to get a residual sequence.
The residual should be linear (or constant). Find its formula.
Add the two parts: $a_{n} = An^{2} + (\text{residual formula})$ .

Worked. Find the nth term of $4, 9, 16, 25, 36, \ldots$ .

1st differences: $5, 7, 9, 11$ . 2nd differences: $2, 2, 2$ — constant!
$A = 1$ . So $An^{2} = n^{2}$ .
Residual: $4 - 1 = 3$ , $9 - 4 = 5$ , $16 - 9 = 7$ , $25 - 16 = 9$ . Sequence: $3, 5, 7, 9$ .
Linear, $d = 2$ . nth term: $2n + 1$ .
Combine: $a_{n} = n^{2} + 2n + 1 = (n + 1)^{2}$ .

Note. This particular sequence is just $(n + 1)^{2}$ — the squares shifted by one.

When the second difference is constant the leading coefficient A is half of it.

Constant 2nd difference = quadratic.
Leading coefficient $A = \dfrac{1}{2}(\text{2nd diff})$ .
Subtract $An^{2}$ to get a linear residual.
Find the linear nth term, add to $An^{2}$ .

Geometric sequences

Constant RATIO between consecutive terms. nth term: $a_{1} \cdot r^{n-1}$ .

A sequence is geometric when each term is the previous term multiplied by a constant ratio $r$ .

Worked. $3, 6, 12, 24, \ldots$ .

$r = 2$ . $a_{1} = 3$ .
$a_{n} = 3 \cdot 2^{n - 1}$ .
$a_{6} = 3 \cdot 32 = 96$ .

Multiplying by the same ratio each step gives aₙ = a₁ · rⁿ⁻¹.

Worked. $80, 40, 20, 10, \ldots$ .

$r = \dfrac{1}{2}$ . $a_{1} = 80$ .
$a_{n} = 80 \cdot \left(\dfrac{1}{2}\right)^{n - 1}$ .

Spotting a geometric sequence. Divide consecutive terms — if you get the same ratio every time, it's geometric.

Real contexts. Compound interest, exponential growth, radioactive decay. We'll meet these formally on the Exponential Growth & Decay notes.

$a_{n} = a_{1} \cdot r^{n - 1}$ .
$r$ = ratio of consecutive terms.
$|r| > 1$ : terms grow. $|r| < 1$ : terms shrink toward zero.
Negative $r$ alternates signs.

Special sequences to recognise

Squares, cubes, triangular numbers, Fibonacci, powers of two — Cambridge expects you to spot these on sight.

Memorise these formulae:

Sequence	First few terms	nth term
Natural numbers	$1, 2, 3, 4, 5, \ldots$	$n$
Even numbers	$2, 4, 6, 8, 10, \ldots$	$2n$
Odd numbers	$1, 3, 5, 7, 9, \ldots$	$2n - 1$
Squares	$1, 4, 9, 16, 25, \ldots$	$n^{2}$
Cubes	$1, 8, 27, 64, 125, \ldots$	$n^{3}$
Triangular	$1, 3, 6, 10, 15, \ldots$	$\dfrac{n(n + 1)}{2}$
Powers of two	$2, 4, 8, 16, 32, \ldots$	$2^{n}$
Fibonacci	$1, 1, 2, 3, 5, 8, 13, \ldots$	$a_{n} = a_{n-1} + a_{n-2}$ (recursive)

Each triangular number adds a new row of dots; the running total is n(n + 1) / 2.

Tip. When you see a sequence on the paper, check: differences? Ratios? Or does it match a famous sequence?

Memorise: squares, cubes, triangular numbers, powers of 2.
Triangular: $\dfrac{n(n+1)}{2}$ .
Fibonacci: each term = sum of previous two.
Recognise on sight to save time.

Quick recap

Linear: 1st diff constant. $a_{n} = dn + c$ .
Quadratic: 2nd diff constant. $A = \dfrac{1}{2}(\text{2nd diff})$ .
Geometric: ratio constant. $a_{n} = a_{1} r^{n - 1}$ .
Squares, cubes, triangular numbers, powers of 2 — memorise.
Substitute $n$ to find term; set formula = value to find $n$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Linear (arithmetic) sequence — constant common difference between consecutive terms.
Geometric sequence — constant common ratio between consecutive terms.
Quadratic sequence — constant second difference.
nth term — a formula for the $n$ -th member of the sequence.
Triangular number — $T_{n} = \dfrac{n(n+1)}{2}$ .

How it’s examined

Sequences appear most years on Paper 2 as 2-3 mark items: continue the sequence, find the nth term. Paper 4 has them as 4-5 mark questions, often combining linear and quadratic, or asking for which term equals a given value. Examiner reports flag the quadratic 'subtract $An^{2}$ ' step as the recurring difficulty.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.13); 0580/22 May/Jun 2024 — Q10 (linear nth term); 0580/42 Oct/Nov 2024 — Q14 (quadratic sequence); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Sequences and nth Term

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

Question type:

Question patterns to master — Sequences and nth Term

Almost every sequences and nth term exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the nth term, find the $k$ th term, which term equals … — on one sequence.

How to approach it

Test the differences: constant first differences mean a linear rule $dn+(a-d)$ ; constant second differences mean a quadratic — the $n^2$ coefficient is half the second difference, with a residual linear sequence to fit.

Common trap

Using the whole second difference as the $n^2$ coefficient instead of half it. Examiner reports flag this, plus writing $T_n=dn$ and dropping the constant correction.

Identify & classify

Recognise it by

Identify the type of sequence or state what kind of sequence this is — square, cube, triangular, geometric.

How to approach it

Compare the terms against the known special sequences ( $n^2$ , $n^3$ , $\frac{n(n+1)}{2}$ , $ar^{n-1}$ ); test the ratio of consecutive terms, not just the difference.

Common trap

Confusing a geometric sequence with a linear one. Examiner reports note candidates check differences only — check the ratio to spot geometric growth.

Multi-step problem

Recognise it by

A non-standard sequence — often a sequence of fractions — where one rule does not describe the whole term.

How to approach it

Split the term into parts (numerator and denominator), find a separate nth-term rule for each sub-pattern, then recombine.

Common trap

Trying to fit a single rule to the whole fraction. The mark scheme credits the numerator rule, denominator rule and combination separately.

1Find the nth term of a linear sequence
CoreDirect calculation• linear sequence
Question
The first four terms of a sequence are $5, 8, 11, 14$ . Find the nth term.
Step-by-step solution
1. Step 1
  Find the common difference.
  $d = 8 - 5 = 3$
2. Step 2
  The nth term is $dn + (a - d)$ where $a$ is the first term.
  $T_{n} = 3n + (5 - 3) = 3n + 2$
3. Step 3
  Verify: $T_{1} = 3(1) + 2 = 5$ ✓
Answer
$T_{n} = 3n + 2$
2Use the nth term to find a specific term
CoreDirect calculation• nth term
Question
The nth term of a sequence is $4n - 7$ . Find the $20$ th term.
Step-by-step solution
1. Step 1
  Substitute $n = 20$ .
  $T_{20} = 4(20) - 7 = 73$
Answer
$73$
3Find the nth term of a quadratic sequence
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q11• quadratic sequence
Question
The first four terms of a sequence are $3, 8, 15, 24$ . Find the nth term.
Step-by-step solution
1. Step 1
  First differences: $5, 7, 9$ → not constant.
2. Step 2
  Second differences: $2, 2$ → constant. So the nth term is quadratic.
3. Step 3
  Coefficient of $n^{2}$ is half the second difference: $\frac{2}{2} = 1$ . So $T_{n}$ contains $n^{2}$ .
4. Step 4
  Subtract $n^{2}$ from each term: $3 - 1 = 2$ , $8 - 4 = 4$ , $15 - 9 = 6$ , $24 - 16 = 8$ . The remainder is $2n$ .
5. Step 5
  Combine.
  $T_{n} = n^{2} + 2n$
Answer
$T_{n} = n^{2} + 2n$
Examiner tip
Half the (constant) second difference gives the coefficient of $n^{2}$ . This shortcut is worth a method mark even before you finish.
4Find which term equals a given value
CoreDirect calculation• nth term
Question
The nth term of a sequence is $5n - 2$ . Which term has the value $98$ ?
Step-by-step solution
1. Step 1
  Set the nth term equal to $98$ and solve.
  $5n - 2 = 98 \implies 5n = 100 \implies n = 20$
Answer
The $20$ th term
5Recognise a special sequence
CoreIdentify & classify• square numbers, cube numbers
Question
Identify the type of sequence: $1, 4, 9, 16, 25, \ldots$ and write its nth term.
Step-by-step solution
1. Step 1
  Each term is a perfect square: $1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, \ldots$
2. Step 2
  nth term:
  $T_{n} = n^{2}$
Answer
Square numbers; $T_{n} = n^{2}$
6Find a term of a quadratic sequence
ExtendedDirect calculation• quadratic sequence, nth term
Question
The nth term of a sequence is $T_{n} = 2n^{2} - n + 3$ . Find the $10$ th term.
Step-by-step solution
1. Step 1
  Substitute $n = 10$ with brackets.
  $T_{10} = 2(10)^{2} - 10 + 3$
2. Step 2
  Evaluate.
  $= 2(100) - 10 + 3 = 200 - 10 + 3 = 193$
Answer
$T_{10} = 193$
7Find the nth term given three terms (quadratic)
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2023 Q13• quadratic sequence, second differences
Question
A quadratic sequence begins $7,\ 14,\ 23,\ 34,\ 47, \ldots$ . Find the nth term.
Step-by-step solution
1. Step 1
  First differences: $7, 9, 11, 13$ — not constant.
2. Step 2
  Second differences: $2, 2, 2$ — constant.
3. Step 3
  Coefficient of $n^{2}$ is half the second difference: $\frac{2}{2} = 1$ . Subtract $n^{2}$ from each term: $7 - 1 = 6$ , $14 - 4 = 10$ , $23 - 9 = 14$ , $34 - 16 = 18$ , $47 - 25 = 22$ .
4. Step 4
  The residual sequence $6, 10, 14, 18, 22$ is linear with common difference $4$ . nth term: $4n + 2$ .
5. Step 5
  Combine.
  $T_{n} = n^{2} + 4n + 2$
Answer
$T_{n} = n^{2} + 4n + 2$
Examiner tip
The mark scheme awards a method mark for the residual-sequence step. The examiner report flags candidates who try simultaneous equations directly and slip on the algebra.
8Find the position of a term in a quadratic sequence
ExtendedDirect calculation• quadratic, solve
Question
The nth term of a sequence is $T_{n} = n^{2} + 3n$ . Find the value of $n$ for which $T_{n} = 130$ .
Step-by-step solution
1. Step 1
  Set up the equation.
  $n^{2} + 3n = 130 \implies n^{2} + 3n - 130 = 0$
2. Step 2
  Factor.
  $(n - 10)(n + 13) = 0 \implies n = 10 \text{ or } n = -13$
3. Step 3
  Reject $n = -13$ because $n$ is a positive integer.
Answer
$n = 10$
9Geometric sequence: find the nth term
ChallengeDirect calculation• geometric sequence
Question
A sequence begins $3,\ 6,\ 12,\ 24,\ 48, \ldots$ . State the type of sequence, find the common ratio, and write down the nth term.
Step-by-step solution
1. Step 1
  Each term is double the previous, so this is a geometric sequence with first term $a = 3$ and common ratio $r = 2$ .
2. Step 2
  Use $T_{n} = ar^{n-1}$ .
  $T_{n} = 3 \times 2^{n - 1}$
3. Step 3
  Check: $T_{1} = 3$ , $T_{2} = 6$ , $T_{3} = 12$ . Confirmed.
Answer
Geometric, $r = 2$ ; $T_{n} = 3 \times 2^{n - 1}$
Examiner tip
The examiner report flags candidates who confuse geometric with linear sequences. Test the RATIO of consecutive terms, not just the difference.
10Find a non-standard pattern
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q16• pattern, synoptic
Question
The sequence is $\dfrac{1}{2},\ \dfrac{2}{5},\ \dfrac{3}{8},\ \dfrac{4}{11},\ \ldots$ . Find the nth term and use it to find $T_{20}$ .
Step-by-step solution
1. Step 1
  Numerators: $1, 2, 3, 4, \ldots$ — the nth numerator is $n$ .
2. Step 2
  Denominators: $2, 5, 8, 11, \ldots$ — linear with common difference $3$ and first term $2$ . nth denominator: $3n - 1$ .
3. Step 3
  Combine.
  $T_{n} = \dfrac{n}{3n - 1}$
4. Step 4
  Substitute $n = 20$ .
  $T_{20} = \dfrac{20}{3(20) - 1} = \dfrac{20}{59}$
Answer
$T_{n} = \dfrac{n}{3n - 1}$ ;\ $T_{20} = \dfrac{20}{59}$
Examiner tip
The mark scheme awards separate marks for finding the numerator rule, the denominator rule and combining. Splitting fractions into two sub-patterns is the recommended approach in the examiner report.

Key Formulae — Sequences and nth Term

The formulae you need to memorise for sequences and nth term on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

nth term of a linear (arithmetic) sequence
$T_{n} = a + (n - 1)d$
$a$
first term
$d$
common difference
$n$
term number, $n \geq 1$
When to use
When the differences between consecutive terms are constant.
Coefficient of $n^{2}$ in a quadratic sequence
$\text{coefficient of } n^{2} = \dfrac{\text{second difference}}{2}$
When to use
When first differences are NOT constant but second differences ARE — sequence is quadratic.

Key Definitions and Keywords — Sequences and nth Term

Definitions to memorise and the exact keywords mark schemes credit for sequences and nth term answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Sequence
Examiner keyword
An ordered list of numbers (terms), each generated by a rule.
Term ( $T_{n}$ )
Examiner keyword
An individual number in a sequence; $T_{n}$ denotes the nth term, where $n$ is the position.
Common difference
Examiner keyword
The fixed amount added to each term to get the next, in a linear/arithmetic sequence.
Quadratic sequence
Examiner keyword
A sequence whose nth term contains $n^{2}$ — recognised by constant second differences.
Square / cube / triangular numbers
Square: $T_{n} = n^{2}$ . Cube: $T_{n} = n^{3}$ . Triangular: $T_{n} = \dfrac{n(n+1)}{2}$ .

Common Mistakes and Misconceptions — Sequences and nth Term

The traps other students keep falling into on sequences and nth term questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Writing $T_{n} = dn$ instead of $T_{n} = dn + (a - d)$
Why it happens
Students stop at the common difference and forget the constant correction.
How to avoid it
After identifying $d$ , find the constant: $T_{1} = a$ tells you what $T_{n} = dn + c$ should give for $n = 1$ .
✕Using the second difference as the coefficient of $n^{2}$ (instead of half)
0580/42 — examiner reports
Why it happens
Easy to forget the divide-by-two step.
How to avoid it
Coefficient of $n^{2}$ = HALF the second difference. Always.
✕Starting the index at $n = 0$
Why it happens
Programming habits leak in.
How to avoid it
In Cambridge IGCSE, $n$ starts at $1$ . The first term is $T_{1}$ , not $T_{0}$ .
✕Not verifying the rule against the given terms
Why it happens
Time pressure.
How to avoid it
Substitute $n = 1, 2, 3$ into your rule and confirm the values match the given sequence.