Sequences · Study Notes · Edexcel IGCSE 4MA1 Higher

Find the $n$ th term of arithmetic, geometric and quadratic sequences. Recognise patterns and produce general formulae.

At a glance

Arithmetic: constant first difference. $u_{n} = a + (n-1)d$ .
Geometric: constant ratio. $u_{n} = ar^{n-1}$ .
Quadratic: constant SECOND difference. Half it for $n^{2}$ coef.
Recursive: each term defined by previous (e.g. Fibonacci).
Verify $n$ th term by testing $n = 1, 2, 3$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.1 — Find the $n$ th term of an arithmetic sequence.
3.1 — Find the $n$ th term of a quadratic sequence.
3.1 — Identify and use geometric sequences.
3.1 — Use recursive (term-to-term) rules.

Arithmetic sequences

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Constant first difference. $n$ th term: $a + (n-1)d$ .

Arithmetic = constant difference between consecutive terms.

Standard form: $u_{n} = a + (n - 1)d$ .

$a$ = first term.
$d$ = common difference.

Equivalent: $u_{n} = dn + (a - d)$ — useful for spotting the formula by inspection.

Example. $5, 8, 11, 14, \ldots$

The same amount is added each step — that constant is the common difference d.

$a = 5$ , $d = 3$ .
$u_{n} = 5 + (n - 1)(3) = 3n + 2$ .
Verify: $n = 1$ : $5$ ✓. $n = 2$ : $8$ ✓.

Spotting by inspection (Edexcel preferred method).

Common difference = coefficient of $n$ .
Adjust constant: subtract that coefficient from the first term to get the constant.
$5, 8, 11, \ldots$ : $d = 3$ , $5 - 3 = 2$ . So $u_{n} = 3n + 2$ .

Worked example. Find the $n$ th term of $40, 35, 30, 25, \ldots$

$d = -5$ .
$40 - (-5) = 45$ .
$u_{n} = -5n + 45$ or $45 - 5n$ .

Edexcel tip. Verify by substituting $n = 1$ and $n = 2$ . Mark scheme credits this verification.

Constant first difference $d$ .
$u_{n} = a + (n-1)d$ .
By inspection: $d \cdot n +$ constant.
Verify $n = 1, 2$ .

Quadratic sequences

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Constant SECOND difference. $n^{2}$ coefficient = half the second difference.

Quadratic sequence = constant SECOND differences (the differences between the differences).

$n$ th term form: $u_{n} = An^{2} + Bn + C$ .

Method:

Find first differences.
Find second differences. Should be constant.
$A =$ half the second difference.
Subtract $An^{2}$ from each term — the result is arithmetic.
Find $B$ (and $C$ ) from the arithmetic part.

Example. $5, 11, 19, 29, 41, \ldots$

Constant second difference 2 confirms a quadratic; the n² coefficient is half of it.

First differences: $6, 8, 10, 12$ .
Second differences: $2, 2, 2$ . ✓ Quadratic.
$A = 2/2 = 1$ . So $n^{2}$ part is $n^{2}$ .
Subtract $n^{2}$ from sequence: $5 - 1 = 4$ , $11 - 4 = 7$ , $19 - 9 = 10$ , $29 - 16 = 13$ , $41 - 25 = 16$ .
$4, 7, 10, 13, 16$ is arithmetic with $d = 3$ , first term $4$ .
$u_{n} = 3n + 1$ for the linear part.
Combined: $u_{n} = n^{2} + 3n + 1$ .

Verify. $n = 1$ : $1 + 3 + 1 = 5$ ✓. $n = 5$ : $25 + 15 + 1 = 41$ ✓.

Worked qualitative. Why does the $n^{2}$ coefficient equal HALF the second difference?

For $u_{n} = An^{2} + Bn + C$ : first differences are $A(2n - 1) + B$ — linear in $n$ .
Second differences are constant $= 2A$ .
So $A =$ half.

Edexcel tip. Always state the second difference and apply the half rule. Mark schemes give M1 for the second difference, M1 for $A$ .

Second difference constant.
$A =$ half the second diff.
Subtract $An^{2}$ , fit arithmetic.
$u_{n} = An^{2} + Bn + C$ .

Geometric sequences

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Constant ratio. $u_{n} = ar^{n-1}$ .

Geometric = constant RATIO between consecutive terms.

Form: $u_{n} = ar^{n-1}$ .

$a$ = first term.
$r$ = common ratio.

Examples:

Sequence	$a$	$r$	$u_{n}$
$2, 6, 18, 54, \ldots$	$2$	$3$	$2 \cdot 3^{n-1}$
$4, 8, 16, 32, \ldots$	$4$	$2$	$4 \cdot 2^{n-1}$
$100, 50, 25, 12.5, \ldots$	$100$	$1/2$	$100 \cdot (1/2)^{n-1}$
$3, -6, 12, -24, \ldots$	$3$	$-2$	$3 \cdot (-2)^{n-1}$

Each term is multiplied by the same ratio r — so the sequence grows multiplicatively.

Worked qualitative. What's the difference between $r > 1$ and $r < 1$ ?

$r > 1$ : terms grow (compound interest, exponential growth).
$0 < r < 1$ : terms shrink (depreciation, decay).
$r$ negative: terms alternate signs.

Edexcel tip. State $a$ and $r$ explicitly. Mark schemes credit each.

$u_{n} = ar^{n-1}$ .
$r$ = common ratio.
$r > 1$ : grow. $0 < r < 1$ : shrink.
Negative $r$ : alternates signs.

Recursive sequences

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Each term defined in terms of previous one(s).

Recursive (term-to-term) rule defines each term using earlier terms.

Standard notation: $u_{1} =$ first term. $u_{n} =$ formula in $u_{n-1}$ (and possibly $u_{n-2}$ ).

Examples:

Arithmetic recursive: $u_{1} = 5$ , $u_{n} = u_{n-1} + 3$ . Equivalent to common difference $3$ .

Geometric recursive: $u_{1} = 2$ , $u_{n} = 3 u_{n-1}$ . Equivalent to common ratio $3$ .

Fibonacci-like: $u_{1} = 1, u_{2} = 1, u_{n} = u_{n-1} + u_{n-2}$ . Each term is the sum of two previous.

Worked example. $u_{1} = 2$ , $u_{2} = 5$ , $u_{n} = u_{n-1} + u_{n-2}$ . Find $u_{5}$ .

$u_{3} = 5 + 2 = 7$ .
$u_{4} = 7 + 5 = 12$ .
$u_{5} = 12 + 7 = 19$ .

Worked qualitative. Why are recursive sequences harder to find general $n$ th terms for?

Each step depends on previous — no closed form in general.
Some have closed forms (Fibonacci has one — too complex for IGCSE).
Edexcel only expects you to GENERATE terms, not find a closed formula.

Edexcel tip. Recursive questions ask you to compute terms one at a time. Show each step.

$u_{n} =$ formula in $u_{n-1}$ .
Generate term-by-term.
Common form: arithmetic, geometric, Fibonacci-style.
No closed formula expected at IGCSE.

Quick recap

Arithmetic: constant first diff, $u_{n} = a + (n-1)d$ .
Geometric: constant ratio, $u_{n} = ar^{n-1}$ .
Quadratic: constant second diff, $u_{n} = An^{2} + Bn + C$ .
$A =$ half the second difference.
Recursive: term-by-term, no closed form expected.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Arithmetic: $u_{n} = a + (n-1)d$ .
Geometric: $u_{n} = ar^{n-1}$ .
Quadratic: $A =$ half second diff.
Verify $n = 1, 2$ always.

How it’s examined

Sequences appear every Paper 1H + 2H (3-7 marks). Arithmetic is foundation; quadratic for higher candidates. Examiner reports flag (1) using full second difference instead of half, (2) wrong sign in arithmetic constant, (3) not verifying.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

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

1 $n$ th term of an arithmetic sequence
Foundation• arithmetic, nth term
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Question
Find the $n$ th term of $5, 8, 11, 14, \ldots$
Step-by-step solution
1. Step 1
  Common difference: $8 - 5 = 3$ . So coefficient of $n$ is $3$ .
2. Step 2
  Test: $3n$ gives $3, 6, 9, 12, \ldots$ . Adjust by adding the difference between $5$ (first term) and $3$ (first $3n$ ): $5 - 3 = 2$ .
3. Step 3
  $n$ th term: $3n + 2$ .
4. Step 4
  Verify: $n = 1$ : $5$ ✓. $n = 4$ : $14$ ✓.
Answer
$3n + 2$
Examiner tip
Always VERIFY by substituting $n = 1$ and $n = 2$ . Mark schemes credit checking.
2Use the $n$ th term to find a specific term
Foundation• Adapted from 4MA1/1H Jan 2024 Q5• nth term
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Question
The $n$ th term of a sequence is $4n - 7$ . Find the 20th term and check whether $73$ is in the sequence.
Step-by-step solution
1. Step 1
  20th term: $4(20) - 7 = 73$ .
2. Step 2
  Set $73 = 4n - 7$ : $4n = 80$ , $n = 20$ . Yes — the 20th term is $73$ .
Answer
20th term = $73$ . $73$ IS in the sequence (term $20$ ).
3 $n$ th term of a quadratic sequence
Higher• Adapted from 4MA1/2H May/Jun 2024 Q14• quadratic sequence
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Question
Find the $n$ th term of $5, 11, 19, 29, 41, \ldots$
Step-by-step solution
1. Step 1
  First differences: $6, 8, 10, 12$ .
2. Step 2
  Second differences: $2, 2, 2$ — constant. Confirms quadratic.
3. Step 3
  Coefficient of $n^{2}$ : half of second difference = $1$ . So formula starts with $n^{2}$ .
4. Step 4
  Subtract $n^{2}$ from sequence: $5 - 1 = 4, 11 - 4 = 7, 19 - 9 = 10, 29 - 16 = 13, 41 - 25 = 16$ . Now $4, 7, 10, 13, 16$ is arithmetic with difference $3$ .
5. Step 5
  $n$ th term of this arithmetic: $3n + 1$ .
6. Step 6
  Combined: $n^{2} + 3n + 1$ .
Answer
$n^{2} + 3n + 1$
Examiner tip
Method: use second differences. Half of second difference = coefficient of $n^{2}$ . Then subtract the $n^{2}$ values, treating remainder as arithmetic.
4Geometric sequence
Higher• geometric
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Question
A geometric sequence has first term $3$ and common ratio $2$ . Find the $n$ th term and the 6th term.
Step-by-step solution
1. Step 1
  Geometric: $a, ar, ar^{2}, ar^{3}, \ldots$ . So $n$ th term: $ar^{n-1}$ .
2. Step 2
  Substitute: $a = 3$ , $r = 2$ . $n$ th term = $3 \cdot 2^{n-1}$ .
3. Step 3
  6th term: $3 \cdot 2^{5} = 3 \cdot 32 = 96$ .
Answer
$n$ th term: $3 \cdot 2^{n-1}$ . 6th term: $96$ .
5Recursive (Fibonacci-style) sequence
Higher• recursive
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Question
A sequence is given by $u_{1} = 2$ , $u_{2} = 5$ , and $u_{n} = u_{n-1} + u_{n-2}$ . Find $u_{5}$ .
Step-by-step solution
1. Step 1
  $u_{3} = u_{2} + u_{1} = 5 + 2 = 7$ .
2. Step 2
  $u_{4} = u_{3} + u_{2} = 7 + 5 = 12$ .
3. Step 3
  $u_{5} = u_{4} + u_{3} = 12 + 7 = 19$ .
Answer
$u_{5} = 19$
6Term-to-term rule through zero [1 mark]
Getting started• Paper 1H/2H style — 1 mark• sequences, term-to-term rule
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Question
Here are the first four terms of a sequence.

$-9, \quad -5, \quad -1, \quad 3, \quad \ldots$

Write down the term-to-term rule for this sequence. [1]
Step-by-step solution
1. Step 1
  Find the change from each term to the next.
  $-5 - (-9) = 4, \quad -1 - (-5) = 4, \quad 3 - (-1) = 4$
2. Step 2
  B1 The change is the same every time, so the rule is to add $4$ .
Answer
Add $4$
Examiner tip
Examiner tip: A term-to-term rule says how to get from one term to the next ("add $4$ "). A position-to-term rule, the $n$ th term, gives a term from its position ( $4n - 13$ here). Answer the one the question asks for.

Common mistake: Writing "subtract $4$ " because the first terms are negative. The numbers are increasing, from $-9$ up to $3$ , so you add.
7nth term when the sequence starts below zero [2 marks]
Getting started• Paper 1H/2H style — 2 marks• sequences, nth term
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Question
Here are the first four terms of an arithmetic sequence.

$-8, \quad -2, \quad 4, \quad 10, \quad \ldots$

Find an expression, in terms of $n$ , for the $n$ th term of the sequence. [2]
Step-by-step solution
1. Step 1
  M1 The common difference is $6$ and the first term is $-8$ . Substitute into $a + (n - 1)d$ .
  $-8 + 6(n - 1)$
2. Step 2
  A1 Expand and simplify.
  $-8 + 6n - 6 = 6n - 14$
Answer
$6n - 14$
Examiner tip
Examiner tip: The June 2024 examiner report notes that an unsimplified answer in the form $a + (n - 1)d$ was credited on an $n$ th-term question. Simplifying is still good practice, because later parts often need the simple form. Check: $n = 3$ gives $18 - 14 = 4$ .

Common mistake: Getting the constant wrong when the first term is negative: $6 \times 1 = 6$ , and $6 - 14 = -8$ , so the constant is $-14$ , not $-2$ or $+14$ .
8Filling in missing terms [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• sequences, common difference, nth term
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Question
The first four terms of an arithmetic sequence are

$7, \quad \square, \quad \square, \quad 22$

(a) Work out the two missing terms. [2]

(b) Write down an expression, in terms of $n$ , for the $n$ th term of the sequence. [1]
Step-by-step solution
1. Step 1
  M1 (a) From the 1st term to the 4th term is $3$ steps of the common difference.
  $3d = 22 - 7 = 15 \quad \Rightarrow \quad d = 5$
2. Step 2
  A1 Add $5$ each time.
  $7, \; 12, \; 17, \; 22$
3. Step 3
  B1 (b) $5n$ gives $5$ for $n = 1$ ; add $2$ to reach $7$ (ft their $d$ ).
  $5n + 2$
Answer
(a) $12$ and $17$ (b) $5n + 2$
Examiner tip
Examiner tip: Count the gaps, not the terms. Four terms have three gaps between them, so divide the total change, $15$ , by $3$ .

Common mistake: Dividing by $4$ and using $d = 3.75$ , or splitting the gap in half ( $14.5$ ). Check your terms: $17 + 5 = 22$ confirms $d = 5$ .
9Sum of the multiples of 7 between 100 and 300 [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• arithmetic series, multiples, sum
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Question
Work out the sum of all the multiples of $7$ between $100$ and $300$ . [4]
Step-by-step solution
1. Step 1
  B1 Find the first and last multiples of $7$ in the range. $7 \times 14 = 98$ is too small and $7 \times 43 = 301$ is too big.
  $7 \times 15 = 105, \qquad 7 \times 42 = 294$
2. Step 2
  M1 The multiples run from $7 \times 15$ to $7 \times 42$ , so count them.
  $n = 42 - 15 + 1 = 28$
3. Step 3
  M1 This is an arithmetic series with $a = 105$ and $d = 7$ . Use the sum formula.
  $S_{28} = \dfrac{28}{2}\left[2 \times 105 + 27 \times 7\right] = 14 \times 399$
4. Step 4
  A1 Evaluate.
  $S_{28} = 5586$
Answer
$5586$
Examiner tip
Examiner tip: Recognise the multiples of a number as an arithmetic sequence whose common difference is that number. Once you know the first term, last term and number of terms, the formula does the rest.

Common mistake: Counting $42 - 15 = 27$ multiples. Both ends are included, so there are $28$ . Using $27$ gives $\dfrac{27}{2}(105 + 287) = 5292$ , which leaves out $294$ .
10Daily running distances [5 marks]
Stretch• Paper 1H/2H style — 5 marks• arithmetic series, context, finding a and d
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Question
Leah goes running every day. The distances she runs each day form an arithmetic sequence.

On day 5 she runs $3.2$ km. On day 13 she runs $5.6$ km.

(a) Work out the distance Leah runs on day 1. [3]

(b) Work out the total distance Leah runs in the first $30$ days. [2]
Step-by-step solution
1. Step 1
  M1 (a) From day 5 to day 13 is $8$ steps of the common difference.
  $8d = 5.6 - 3.2 = 2.4$
2. Step 2
  M1 $d = 0.3$ . Go back $4$ steps from day 5.
  $a = 3.2 - 4 \times 0.3$
3. Step 3
  A1 Evaluate.
  $a = 2 \text{ km}$
4. Step 4
  M1 (b) Use the sum formula with $a = 2$ , $d = 0.3$ and $n = 30$ (ft their $a$ and $d$ ).
  $S_{30} = \dfrac{30}{2}\left[2 \times 2 + 29 \times 0.3\right] = 15 \times 12.7$
5. Step 5
  A1 Evaluate.
  $S_{30} = 190.5 \text{ km}$
Answer
(a) $2$ km (b) $190.5$ km
Examiner tip
Examiner tip: Check part (a) against the other given day: $2 + 12 \times 0.3 = 5.6$ km on day 13, as stated.

Common mistake: In part (b), giving the distance run on day 30 ( $2 + 29 \times 0.3 = 10.7$ km) instead of the total. Read the question: "total distance" means a sum, not a single term.

Question 1
Paper 1H/2H style — 1 mark1 mark
Here are the first five terms of a sequence.

$1, \quad 4, \quad 9, \quad 16, \quad 25, \quad \ldots$

Write down the next two terms of the sequence. [1]
Model answer
These are the square numbers $1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}$ .

Next two terms: $6^{2} = 36$ and $7^{2} = 49$
Why this scores
Mark scheme:
- B1 for $36$ and $49$
Why this scores full marks: The candidate recognises the square numbers instead of looking for a common difference. The differences $3, 5, 7, 9$ go up by $2$ , which gives the same two terms: $25 + 11 = 36$ and $36 + 13 = 49$ .
Question 2
Paper 1H/2H style — 1 mark1 mark
The term-to-term rule of a sequence is: subtract $8$

The 3rd term of the sequence is $5$ .

Write down the 1st term of the sequence. [1]
Model answer
Work backwards with the inverse of the rule, add $8$ :

2nd term $= 5 + 8 = 13$

1st term $= 13 + 8 = 21$

Answer: $21$
Why this scores
Mark scheme:
- B1 for $21$ cao
Why this scores full marks: The candidate goes backwards from the 3rd term, so the inverse operation, add $8$ , is used twice. Check forwards: $21, 13, 5$ each subtract $8$ . Applying the rule itself backwards gives $5 - 8 - 8 = -11$ , which is the 5th term, not the 1st.
Question 3
Paper 1H/2H style — 2 marks2 marks
The first term of a sequence is $4$ .

The term-to-term rule of the sequence is: multiply by $3$ then subtract $5$

Work out the 4th term of the sequence. [2]
Model answer
2nd term: $4 \times 3 - 5 = 7$

3rd term: $7 \times 3 - 5 = 16$

4th term: $16 \times 3 - 5 = 43$

Answer: $43$
Why this scores
Mark scheme:
- M1 for $7$ and $16$ seen
- A1 for $43$ cao
Why this scores full marks: The candidate applies the rule one term at a time and writes each term down, so the method mark is secure. The order matters: multiply first, then subtract. Doing it the other way round gives $(4 - 5) \times 3 = -3$ and the wrong sequence.
Question 4
Paper 1H/2H style — 2 marks2 marks
The $n$ th term of sequence A is $3n + 1$ .

The $n$ th term of sequence B is $5n - 2$ .

Find the two smallest numbers that are terms of both sequence A and sequence B. [2]
Model answer
Sequence A: $4, 7, 10, 13, 16, 19, 22, 25, 28, \ldots$

Sequence B: $3, 8, 13, 18, 23, 28, \ldots$

Answer: $13$ and $28$
Why this scores
Mark scheme:
- M1 for at least the first five terms of both sequences correct, or $13$ identified as a common term
- A1 for $13$ and $28$ only
Why this scores full marks: The candidate generates terms from each position-to-term rule, starting at $n = 1$ , and lists far enough to find two matches. The common terms go up by $15$ , the lowest common multiple of the two differences $3$ and $5$ , so $28 = 13 + 15$ is a quick check.
Question 5
Paper 1H/2H style — 3 marks3 marks
The first term of an arithmetic series is $-7$ . The common difference is $2.5$

(a) Work out the 20th term of the series. [1]

(b) Work out the sum of the first $20$ terms of the series. [2]
Model answer
(a) $-7 + 19 \times 2.5 = -7 + 47.5 = 40.5$

(b) $S_{20} = \dfrac{20}{2}\left[2 \times (-7) + 19 \times 2.5\right] = 10(-14 + 47.5) = 10 \times 33.5$

Answer: $335$
Why this scores
Mark scheme:
- B1 for $40.5$ (a)
- M1 for $\dfrac{20}{2}\left[2 \times (-7) + 19 \times 2.5\right]$ oe, or $\dfrac{20}{2}(-7 + 40.5)$ (b)
- A1 for $335$ cao (b)
Why this scores full marks: The candidate puts $-7$ in brackets, so $2 \times (-7) = -14$ is not lost. In (b) either form of the sum earns the M1; using the answer to (a) as the last term gives $10 \times 33.5 = 335$ as a check.
Question 6
Paper 1H/2H style — 3 marks3 marks
Use the formula for the sum of an arithmetic series to prove that the sum of the first $n$ odd numbers is $n^{2}$ [3]
Model answer
The odd numbers $1, 3, 5, 7, \ldots$ form an arithmetic series with $a = 1$ and $d = 2$ .

$S_{n} = \dfrac{n}{2}\left[2 \times 1 + (n - 1) \times 2\right]$

$= \dfrac{n}{2}(2 + 2n - 2) = \dfrac{n}{2} \times 2n = n^{2}$

So the sum of the first $n$ odd numbers is $n^{2}$ .
Why this scores
Mark scheme:
- B1 for $a = 1$ and $d = 2$ stated or used
- M1 for $\dfrac{n}{2}\left[2 + 2(n - 1)\right]$ oe
- A1 for a complete, correct simplification to $n^{2}$ with a conclusion
Why this scores full marks: Checking a few cases ( $1 + 3 = 4$ , $1 + 3 + 5 = 9$ ) is not a proof. The candidate uses the general formula, simplifies the bracket fully and ends with a statement that links back to the question.
Question 7
Paper 1H/2H style — 4 marks4 marks
The first three terms of an arithmetic sequence are

$2x + 1, \quad 3x + 2, \quad 5x - 1$

(a) Find the value of $x$ . [2]

(b) Find an expression, in terms of $n$ , for the $n$ th term of the sequence. [2]
Model answer
(a) The common difference is the same between each pair of terms.

$(3x + 2) - (2x + 1) = (5x - 1) - (3x + 2)$

$x + 1 = 2x - 3 \quad \Rightarrow \quad x = 4$

(b) The terms are $9, 14, 19$ , so $d = 5$ .

$9 + 5(n - 1) = 5n + 4$

Answer: $5n + 4$
Why this scores
Mark scheme:
- M1 for $(3x + 2) - (2x + 1) = (5x - 1) - (3x + 2)$ oe (a)
- A1 for $x = 4$ (a)
- M1 for finding the terms $9, 14, 19$ or $d = 5$ , ft their $x$ (b)
- A1 for $5n + 4$ oe (b)
Why this scores full marks: The candidate uses the defining property of an arithmetic sequence: equal differences. Brackets round $(3x + 2)$ make sure both terms are subtracted, which avoids the sign error $5x - 1 - 3x + 2$ . Substituting $x = 4$ back gives three terms with difference $5$ , which checks part (a).
Question 8
Paper 1H/2H style — 4 marks4 marks
Here is an arithmetic series.

$5 + 13 + 21 + 29 + \ldots$

(a) Show that the sum of the first $n$ terms of the series is $n(4n + 1)$ [2]

(b) Work out the sum of the 11th term to the 20th term inclusive. [2]
Model answer
(a) $a = 5$ , $d = 8$

$S_{n} = \dfrac{n}{2}\left[10 + 8(n - 1)\right] = \dfrac{n}{2}(8n + 2) = n(4n + 1)$

(b) $S_{20} = 20 \times 81 = 1620$ and $S_{10} = 10 \times 41 = 410$

$S_{20} - S_{10} = 1620 - 410$

Answer: $1210$
Why this scores
Mark scheme:
- M1 for $\dfrac{n}{2}\left[2 \times 5 + (n - 1) \times 8\right]$ oe (a)
- A1 for a complete, correct simplification to $n(4n + 1)$ (a)
- M1 for $S_{20} - S_{10}$ oe, e.g. $1620 - 410$ , or $\dfrac{10}{2}(85 + 157)$ (b)
- A1 for $1210$ cao (b)
Why this scores full marks: Terms 11 to 20 are the first 20 terms with the first 10 taken away, so the candidate subtracts $S_{10}$ , not $S_{11}$ . Subtracting $S_{11} = 495$ would also remove the 11th term and give $1125$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
Omar saves £ $12$ in the first month. Each month after that, he saves £ $4$ more than in the month before.

(a) Show that the total amount, in pounds, Omar has saved after $n$ months is $2n(n + 5)$ [2]

(b) Work out the number of months it takes Omar to save a total of £ $1000$ . Show clear algebraic working. [3]
Model answer
(a) $a = 12$ , $d = 4$

$S_{n} = \dfrac{n}{2}\left[24 + 4(n - 1)\right] = \dfrac{n}{2}(4n + 20) = n(2n + 10) = 2n(n + 5)$

(b) $2n(n + 5) = 1000 \quad \Rightarrow \quad n^{2} + 5n - 500 = 0$

$(n + 25)(n - 20) = 0 \quad \Rightarrow \quad n = -25 \text{ or } n = 20$

$n$ is a number of months, so it cannot be negative.

Answer: $20$ months
Why this scores
Mark scheme:
- M1 for $\dfrac{n}{2}\left[2 \times 12 + (n - 1) \times 4\right]$ oe (a)
- A1 for a complete, correct simplification to $2n(n + 5)$ (a)
- M1 for $n^{2} + 5n - 500 = 0$ oe, e.g. $2n^{2} + 10n - 1000 = 0$ (b)
- M1 for $(n + 25)(n - 20)$ , or a correct method to solve their 3-term quadratic (b)
- A1 for $20$ only, with $-25$ rejected (b)
Why this scores full marks: The candidate divides by $2$ first, which makes the quadratic easy to factorise, and states why $-25$ is rejected. Check: in month $20$ Omar saves $12 + 19 \times 4 = 88$ pounds, and $\dfrac{20}{2}(12 + 88) = 1000$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
Series A is $3 + 7 + 11 + 15 + \ldots$

Series B is $40 + 42 + 44 + 46 + \ldots$

The sum of the first $n$ terms of series A is equal to the sum of the first $n$ terms of series B.

Find the value of $n$ . Show clear algebraic working. [5]
Model answer
Series A: $a = 3$ , $d = 4$

$S_{n} = \dfrac{n}{2}\left[6 + 4(n - 1)\right] = \dfrac{n}{2}(4n + 2) = n(2n + 1)$

Series B: $a = 40$ , $d = 2$

$S_{n} = \dfrac{n}{2}\left[80 + 2(n - 1)\right] = \dfrac{n}{2}(2n + 78) = n(n + 39)$

$n(2n + 1) = n(n + 39) \quad \Rightarrow \quad 2n^{2} + n = n^{2} + 39n$

$n^{2} - 38n = 0 \quad \Rightarrow \quad n(n - 38) = 0$

$n = 0$ is not possible (there must be at least one term), so

Answer: $n = 38$
Why this scores
Mark scheme:
- M1 for $\dfrac{n}{2}\left[2 \times 3 + (n - 1) \times 4\right]$ oe (series A)
- M1 for $\dfrac{n}{2}\left[2 \times 40 + (n - 1) \times 2\right]$ oe (series B)
- A1 for $n(2n + 1)$ and $n(n + 39)$ oe
- M1 for equating their sums and reducing to $n^{2} - 38n = 0$ oe
- A1 for $38$ only
Why this scores full marks: The candidate rearranges to $= 0$ and factorises, so the solution $n = 0$ appears and is rejected with a reason. Dividing both sides by $n$ also works, but only because $n \neq 0$ , and a strong script says so. Check: both sums equal $38 \times 77 = 2926$ .

4MA1 Sequences

At a glance

What you’ll learn

Arithmetic sequences

Quadratic sequences

Geometric sequences

Recursive sequences

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1 $n$ th term of an arithmetic sequence

2Use the $n$ th term to find a specific term

3 $n$ th term of a quadratic sequence

4Geometric sequence

5Recursive (Fibonacci-style) sequence

6Term-to-term rule through zero [1 mark]

7nth term when the sequence starts below zero [2 marks]

8Filling in missing terms [3 marks]

9Sum of the multiples of 7 between 100 and 300 [4 marks]

10Daily running distances [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Sequences

At a glance

What you’ll learn

Arithmetic sequences

Quadratic sequences

Geometric sequences

Recursive sequences

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1nnnth term of an arithmetic sequence

2Use the nnnth term to find a specific term

3nnnth term of a quadratic sequence

4Geometric sequence

5Recursive (Fibonacci-style) sequence

6Term-to-term rule through zero [1 mark]

7nth term when the sequence starts below zero [2 marks]

8Filling in missing terms [3 marks]

9Sum of the multiples of 7 between 100 and 300 [4 marks]

10Daily running distances [5 marks]

Model AnswersFull mark-scheme-style answers

1 $n$ th term of an arithmetic sequence

2Use the $n$ th term to find a specific term

3 $n$ th term of a quadratic sequence