Edexcel IGCSE

Sequences

A sequence is a list of numbers in a fixed order, and the questions are about predicting it. It is section 3.1 of the Edexcel International GCSE Mathematics A specification, and almost every question on it reduces to the same move: find how much the list goes up by each time, then use that step to jump straight to any term you want without writing the list out.

Edexcel IGCSE 4MA1Specification 3.319 approved practice questions

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Before you start

What you need to be able to do already

If any of these is shaky, fix it first — every worked example below leans on all three.

Add and subtract with negative numbers, and keep the sign
Work out $14 - 17$ , and then $20 + (- 3)$ . They are $- 3$ and $17$ . A decreasing sequence has a negative common difference, so both of those moves — subtracting to find the difference, then adding it back on — appear in every step of the work below.
Expand a bracket and collect like termsAlgebraic manipulation
Simplify $7 + 4 (n - 1)$ . It is $4 n + 3$ : expand to $7 + 4 n - 4$ , then collect. That single line is how the Higher-tier formula and the Foundation-tier answer turn out to be the same expression, so the two halves of this topic only join up if you can do it.
Substitute a number into a formula and rearrange to find an unknown
If $T = a + 4 d$ , $T = 23$ and $a = 7$ , find $d$ . It is $4$ : subtract $7$ to get $4 d = 16$ , then divide. Higher-tier questions hand you two terms and ask for $a$ and $d$ , which is this move done twice.

The idea

What is sequences?

A sequence is an ordered list of numbers, and each number in it is a term. There are two ways to describe one. A term-to-term rule says how to get from any term to the next, such as ‘add $4$ ’; a position-to-term rule, called the nth term, is a formula that turns a position number straight into the term sitting there. A sequence is arithmetic when the step from each term to the next is the same every time, and that constant step is what makes a single formula possible at all.

The notation, and how to read it

Every worked solution below is written in these five symbols.

SymbolReads asMeans
$n$ the position numberWhich term you are talking about, counting from $n = 1$ for the first term. It is always a positive whole number, and it is never a term value.
$a$ the first termThe term at position $n = 1$ . In $7, 11, 15, \dots$ the first term is $a = 7$ .
$d$ the common differenceThe constant step: any term minus the one before it. It is negative when the sequence decreases, so $20, 17, 14, \dots$ has $d = - 3$ .
$a + (n - 1) d$ the nth term of an arithmetic sequenceStart at the first term and take $n - 1$ steps of size $d$ — not $n$ steps, because you are already standing on term $1$ .
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ the sum of the first n termsThe total of the first $n$ terms of an arithmetic series, not a single term. Higher tier only.

The one idea underneath all of it

An arithmetic sequence changes by the same amount at every step, so getting from the first term to the nth term is exactly $n - 1$ steps of size $d$ . That one sentence is the whole formula, $a + (n - 1) d$ , and it also explains its most-missed feature: the $n - 1$ is not a quirk of the algebra, it is the fact that you have already arrived at term $1$ before taking any step at all. Using $n$ where the formula wants $n - 1$ is the error our question bank records most often on this subtopic.

The smallest possible example

Find the nth term of $7, 11, 15, 19, \dots$

$4 n + 3$

The list goes up by $4$ each time, so the formula starts $4 n$ . At $n = 1$ , $4 n$ gives $4$ , which is $3$ short of the first term, so add $3$ . Check at $n = 4$ : $16 + 3 = 19$ , the fourth term. Every other question on this page is this, with the numbers hidden further away.

Key terms

The words a question will use

Exam wording is precise. These five carry the meaning a question depends on.

TermWhat it meansNotation
Term
One number of the sequence, sitting at one position. ‘The 5th term’ means the number in position $5$ , and the first term is always at $n = 1$ , never at $n = 0$ . $n = 1$
Term-to-term rule
How to get from one term to the next, such as ‘add $4$ ’ or ‘multiply by $3$ ’. It is quick for the next term or two and useless for the $100$ th, because you would have to work out all $99$ before it. $+ 4$
Position-to-term rule
The nth term: a formula in $n$ that turns a position straight into its term. Substituting $n = 100$ into $4 n + 3$ gives the hundredth term, $403$ , in a single line. $4 n + 3$
Common difference
The constant amount an arithmetic sequence changes by. Work it out as a term minus the one before it — later minus earlier — or a decreasing sequence comes out with the wrong sign, which our bank records as its own misconception. $d = 11 - 7$
Arithmetic series
The sum of the terms of an arithmetic sequence, rather than the list itself. Higher tier only, and the word ‘series’ or ‘sum’ is the signal that a total is wanted and the nth-term formula is the wrong tool. $S_{n}$

The method

How to work out a sequences

Five steps, in this order. The worked examples below point back to these numbers.

Check that the difference between consecutive terms is the same every time.For $3, 7, 11, 15$ the gaps are $+ 4, + 4, + 4$ , so the sequence is arithmetic and everything below applies. If the gaps change, nothing that follows is valid — say so rather than forcing a linear nth term onto a sequence that does not have one.
Write down the first term $a$ and the common difference $d$ , with its sign.For $20, 17, 14, \dots$ that is $a = 20$ and $d = - 3$ , not $+ 3$ . Take later minus earlier every time. Swapping $a$ and $d$ over, and losing the minus sign on a decreasing sequence, are two separate misconceptions catalogued for this subtopic.
For the nth term, multiply $n$ by $d$ , then adjust so that $n = 1$ gives the first term.For $7, 11, 15, \dots$ start with $4 n$ ; at $n = 1$ that is $4$ , three short of $7$ , so the nth term is $4 n + 3$ . On Higher the same answer comes from $a + (n - 1) d = 7 + 4 (n - 1)$ , which expands to $4 n + 3$ — one formula, two routes.
To find a particular term, substitute its position number for $n$ . $4 n + 3$ at $n = 10$ gives $43$ . With $a + (n - 1) d$ , the 8th term needs $n - 1 = 7$ , so it is $a + 7 d$ and not $a + 8 d$ . This is the off-by-one named above, arriving exactly where it always does, so write $n - 1$ out as a number before you evaluate anything.
Check your formula against a term you already know.Put $n = 1$ and $n = 2$ back into your nth term and confirm you get the first two terms of the list. It takes ten seconds and catches both the off-by-one and a lost minus sign, which between them account for most of the marks dropped here.

Use it when

The question gives a list of numbers, or a first term and a common difference, and asks for the next term, a particular term, the nth term, or on Higher tier a sum. Check the differences are constant first — that is what licenses every step.

Do not use it when

The differences are not constant. $2, 4, 8, 16, \dots$ doubles, so its step is a multiplication and $a + (n - 1) d$ does not apply; sequences whose *second* differences are constant are quadratic and sit outside this section. Also do not reach for the nth-term formula when the word is ‘sum’ or ‘series’: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ is a different formula, and reaching for each of these two in place of the other is recorded as two separate misconceptions in our bank for this subtopic.

Worked examples

6 IGCSE Maths practice questions on sequences, fully worked

One per objective, so the range is covered rather than the same question repeated. Every step is written out, and the answer is shown — this is study material, not a test.

Example 1

The first term of a sequence is 5. The term-to-term rule is 'add 4'. What is the 4th term of this sequence? [1 mark]

Generate terms of a sequence using term-to-term and position-to-term definitions of the sequence

Worked solution

Step 1: The first term is 5.
Step 2: According to the rule, we add 4 to get the next term. So the second term is $5 + 4 = 9$ .
Step 3: To find the third term, add 4 to the second term: $9 + 4 = 13$ .
Step 4: To find the fourth term, add 4 to the third term: $13 + 4 = 17$ .

Answer17

Example 2

A sequence starts 3, 8, 13, 18, … Find the next term in the sequence. [1 mark]

Find subsequent terms of an integer sequence and the rule for generating it

Worked solution

Step 1: Look at the difference between each pair of terms: 8 − $3 = 5$ , 13 − $8 = 5$ , 18 − $13 = 5$ .
Step 2: The common difference is 5.
Step 3: To find the next term, add 5 to the last term: $18 + 5 = 23$ .

Answer23

Example 3

The n-th term of a linear sequence is given by 2n + 5. Work out the 8th term of this sequence. [2 marks]

Use linear expressions to describe the nth term of arithmetic sequences

Worked solution

Step 1: The n-th term formula is 2n + 5. We need the 8th term, so we substitute n with 8.
Step 2: Multiply 2 by 8: $2 \times 8 = 16$ .
Step 3: Add 5 to the result: $16 + 5 = 21$ .
Step 4: The 8th term is 21.

Answer21

Show 3 more worked examples

Example 4

An arithmetic sequence has first term a = 7 and common difference d = 4. Which of the following is the 5th term of the sequence? [2 marks]

Understand and use common difference (d) and first term (a) in an arithmetic sequence

Worked solution

Step 1: The nth term of an arithmetic sequence is given by: nth term = a + (n-1)d.
Step 2: For the 5th term, n = 5. Substitute a = 7 and d = 4.
Step 3: 5th term = $7 + (5 - 1) \times 4 = 7 + 4 \times 4 = 7 + 16 = 23$ .

Answer23

Example 5

The nth term of a sequence is given by $a_{n} = a + (n - 1) d$ . The 5th term is 17 and the 9th term is 29. Find the value of the 3rd term. [4 marks]

Know and use nth term = a + (n - 1)d

Worked solution

Step 1: Write the 5th term: a + 4d = 17.
Step 2: Write the 9th term: a + 8d = 29.
Step 3: Subtract the first equation from the second: (a+8d)-(a+4d)= $29 - 17$ gives 4d=12, so d=3.
Step 4: Substitute d=3 into a+4(3)=17 gives a+ $12 = 17$ , so a=5.
Step 5: Now compute the 3rd term: a + 2d = $5 + 2 (3) = 11$ .

Answer11

Example 6

The sum of the first n terms of an arithmetic series is given by $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . For a particular series, $S_{10} = 230$ and the 10th term is 32. Show that the first term is 14 and find the common difference. Give your answer as an integer, e.g. 17. [4 marks]

Find the sum of the first n terms of an arithmetic series (Sn)

Worked solution

Step 1: Use the nth term formula: 10th term = a + 9d = 32.
Step 2: Use the sum formula for n=10: S10 = 10/2 × (2a + 9d) = 230, so 5(2a + 9d) = 230.
Step 3: Simplify 5(2a + 9d) = 230 to 2a + 9d = 46.
Step 4: Solve the simultaneous equations: a + 9d = 32 and 2a + 9d = 46.
Step 5: Subtract first from second: a = 14. Then substitute back to get $14 + 9$ d = 32, so 9d = 18, d = 2.

Answerd = 2

Common mistakes

Where marks actually get lost

Learnly tags 12 distinct misconceptions on this subtopic. These are errors that show up in real student work — not generic revision advice.

Examiner traps

Using n instead of n-1
What goes wrong: Students often use the formula a + n*d instead of a + (n-1)*d, leading to an off-by-one error in the term number.
What to do instead: Remember: the first term uses d zero times. For the nth term, add d (n-1) times. Formula: a + (n-1)d.
Sign error with negative d
What goes wrong: When the common difference is negative, students may add instead of subtract, or misapply the sign in calculations.
What to do instead: If d is negative, you subtract each time. For example, a=10, d=-3: 2nd term = $10 - 3 = 7$ , 3rd term = $7 - 3 = 4$ .
Confusing first term and common difference
What goes wrong: Students swap the values of a and d, e.g., using a as the common difference and d as the first term.
What to do instead: The first term a is the starting number. The common difference d is what you add each time. Label them clearly.
Arithmetic calculation error
What goes wrong: Students make simple addition or multiplication mistakes when computing the nth term, especially with larger numbers.
What to do instead: Break the calculation into steps: first find (n-1)*d, then add a. Double-check your arithmetic.
Off-by-one error in term index
What goes wrong: Students incorrectly substitute n = 8 into the formula but forget that the first term corresponds to n = 1, leading to using n = 7 for the 8th term.
What to do instead: For the 8th term, n = 8, not 7. Always use the term number directly as n.
Using incorrect nth term formula
What goes wrong: Students mistakenly use the formula for the sum of an arithmetic series or confuse it with other sequences like quadratic.
What to do instead: The nth term of an arithmetic sequence is a + (n-1)d. Do not use sum formulas.

See 6 more examiner traps

Sign error in common difference
What goes wrong: Students misidentify the common difference d, especially when terms are decreasing, leading to incorrect substitution.
What to do instead: Check if the sequence increases or decreases. d can be negative; use it correctly in the formula.
Order of operations mistake
What goes wrong: Students incorrectly compute (n-1)d by multiplying d first and then subtracting, or they forget to multiply d by (n-1) before adding a.
What to do instead: Compute (n-1) first, then multiply by d, then add a. Use brackets if needed.
Using wrong formula
What goes wrong: Students use the formula for the nth term (a + (n-1)d) instead of the sum formula.
What to do instead: Remember: Sum uses Sn = n/2 [2a + (n-1)d]. Don't confuse with nth term formula.
Misplacing n in formula
What goes wrong: Students incorrectly place n in the formula, e.g., writing n/2 * [a + (n-1)d] instead of n/2 * [2a + (n-1)d].
What to do instead: Check the sum formula: Sn = n/2 [2a + (n-1)d]. The '2a' is key, not just 'a'.
Order of operations error
What goes wrong: Students incorrectly compute the expression inside brackets, e.g., adding before multiplying.
What to do instead: First compute (n-1)d, then add 2a, then multiply by n/2. Follow BIDMAS.
Forgetting to multiply by n/2
What goes wrong: Students compute 2a + (n-1)d correctly but forget to multiply by n/2.
What to do instead: After finding the bracket, don't forget to multiply by n/2 to get the sum.

Exam technique

How the marks are actually awarded

Method marks are given for working a marker can follow. This is what to put on the page.

Where the marks are: Questions in this section run from $1$ to $4$ marks. ‘Write down the next term’ is $1$ . Finding the nth term is normally $2$ : one mark for the $4 n$ part, which is the letter $n$ multiplied by the common difference, and one for the constant that corrects it — so write $4 n$ down even if the adjustment then goes wrong, because that mark is already banked. On Higher, being given two terms and asked for $a$ and $d$ is $3$ or $4$ marks, and they sit on the two equations you form rather than on the final numbers, so show both. Sum questions are $3$ to $4$ , with a mark for quoting $S_{n}$ correctly before any substitution.
Command words to expect: Write down the next termFind an expression for the nth termGive your answer in terms of nFind the common differenceFind the first termWork out the value ofShow thatExplain whyFind the sum of the first n terms
Accuracy and rounding: Nothing rounds — every answer here is exact, and usually a whole number or a simple expression. Form carries the marks instead. ‘In terms of $n$ ’ means the answer must contain the letter $n$ , so $4 n + 3$ scores and $43$ does not. Write $4 n + 3$ , not $n 4 + 3$ . A decreasing sequence keeps its minus sign: the nth term of $20, 17, 14, \dots$ is $23 - 3 n$ , and $23 + 3 n$ describes a different sequence entirely. When asked whether a number appears in a sequence, set the nth term equal to it and solve — $4 n + 3 = 43$ gives $n = 10$ , so yes; an answer for $n$ that is a fraction, zero or negative means no.
Calculator: A calculator is allowed in every paper of this qualification, on both Foundation and Higher tiers. It will evaluate $a + (n - 1) d$ faultlessly once you have typed it, which is exactly the part that was never the difficulty — it cannot tell you that the 8th term needs $n - 1 = 7$ . Use it for the arithmetic inside a sum question, and settle the formula by hand first.

Check yourself

You should now be able to:

Continue a sequence from a term-to-term rule such as ‘add $4$ ’, and state the rule when only the list is given.
Generate the terms of a sequence from a position-to-term rule by substituting $n = 1, 2, 3$ .
Show that a sequence is arithmetic by checking that consecutive differences are equal.
Find the common difference as later term minus earlier term, keeping the minus sign when the sequence decreases.
Write the nth term of an arithmetic sequence as a linear expression in $n$ , such as $4 n + 3$ .
Find any term of a sequence by substituting its position number into the nth term.
Decide whether a given number is in a sequence by solving for $n$ and checking it is a positive whole number.
Use $a$ for the first term and $d$ for the common difference, and apply $a + (n - 1) d$ (Higher).
Find $a$ and $d$ from two given terms by forming and solving a pair of equations (Higher).
Find the sum of the first $n$ terms with $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , and tell a sum question from a term question (Higher).

Specification coverage

The 6 things 4MA1 asks you to do

The assessable objectives for sequences, functions and graphs — sequences, in the order the specification lists them.

Generate terms of a sequence using term-to-term and position-to-term definitions of the sequence
Find subsequent terms of an integer sequence and the rule for generating it
Use linear expressions to describe the nth term of arithmetic sequences
Understand and use common difference (d) and first term (a) in an arithmetic sequence
Know and use nth term = a + (n - 1)d
Find the sum of the first n terms of an arithmetic series (Sn)

See who is stuck on sequences before you teach it.

Create a class, share the 8-character code, and set sequences as practice. Your students work through all 19 approved practice questions — not just the 6 on this page — and you see which objectives the class has not got yet. Free for individual students and teachers.

How Learnly works for teachers

Set up your class — free Practise on your own

What you need to be able to do already

Add and subtract with negative numbers, and keep the sign

Expand a bracket and collect like termsAlgebraic manipulation

Substitute a number into a formula and rearrange to find an unknown

What is sequences?

The notation, and how to read it

The words a question will use

Term

Term-to-term rule

Position-to-term rule

Common difference

Arithmetic series

How to work out a sequences

6 IGCSE Maths practice questions on sequences, fully worked

The first term of a sequence is 5. The term-to-term rule is 'add 4'. What is the 4th term of this sequence? [1 mark]

A sequence starts 3, 8, 13, 18, … Find the next term in the sequence. [1 mark]

The n-th term of a linear sequence is given by 2n + 5. Work out the 8th term of this sequence. [2 marks]

An arithmetic sequence has first term a = 7 and common difference d = 4. Which of the following is the 5th term of the sequence? [2 marks]

The nth term of a sequence is given by an​=a+(n−1)d. The 5th term is 17 and the 9th term is 29. Find the value of the 3rd term. [4 marks]

The sum of the first n terms of an arithmetic series is given by Sn​=2n​[2a+(n−1)d]. For a particular series, S10​=230 and the 10th term is 32. Show that the first term is 14 and find the common difference. Give your answer as an integer, e.g. 17. [4 marks]

Where marks actually get lost

Using n instead of n-1

Sign error with negative d

Confusing first term and common difference

Arithmetic calculation error

Off-by-one error in term index

Using incorrect nth term formula

Sign error in common difference

Order of operations mistake

Using wrong formula

Misplacing n in formula

Order of operations error

Forgetting to multiply by n/2

How the marks are actually awarded

You should now be able to:

The 6 things 4MA1 asks you to do

See who is stuck on sequences before you teach it.

The nth term of a sequence is given by $a_{n} = a + (n - 1) d$ . The 5th term is 17 and the 9th term is 29. Find the value of the 3rd term. [4 marks]

The sum of the first n terms of an arithmetic series is given by $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . For a particular series, $S_{10} = 230$ and the 10th term is 32. Show that the first term is 14 and find the common difference. Give your answer as an integer, e.g. 17. [4 marks]