What is 4.2: Sequences?

An explicit rule gives x_n directly from n, whereas a recurrence relation defines each term from one or more preceding terms and needs an initial value.

How many marks are 4.2 questions worth?

4.2 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 4.2 appeared in?

We have not yet indexed a verified past-paper appearance for 4.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

4.2

Work with sequences including those given by a formula for the nth term and those generated by a simple relation of the form xₙ₊₁ = f(xₙ); increasing sequences; decreasing sequences; periodic sequences.

Draft — not yet indexed

Sequences

Worked answers and methods for 4.2 on Edexcel A-level Maths 9MA0.

Explanation

An explicit rule gives $x_n$ directly from $n$ , whereas a recurrence relation defines each term from one or more preceding terms and needs an initial value.
A sequence is increasing when $x_{n+1}>x_n$ and decreasing when $x_{n+1}<x_n$ throughout the stated range of $n$ .
A sequence is periodic if its terms repeat after a fixed positive number of steps; the least such number is its period.
When using a recurrence, retain sufficient accuracy between steps and check any invariant interval.
A common error is to apply $f$ repeatedly to the initial term instead of to the latest term.

Worked example

A sequence is defined by $x_1=1$ and $x_{n+1}=5-x_n$ . Find $x_2,x_3,x_4,x_5$ and state its period.

1.Apply the recurrence to the most recent term: $x_2=5-1=4$ , $x_3=5-4=1$ , and the pair $1,4$ then repeats.
2.Thus $x_4=4$ , $x_5=1$ , and the least period is $2$ .

Answer: $x_2=4$ , $x_3=1$ , $x_4=4$ , $x_5=1$ ; Period $2$

Common mistakes

Don't declare a sequence increasing because $x_2>x_1$ without checking later terms or the recurrence.
Don't classify a recurrence from too few terms and miss a repeating cycle.

Exam tip

Generate enough consecutive terms to demonstrate the claimed monotonicity or period rather than guessing from the first change.

Worked practice

Tier 1 · Easy

The sequence

(x_n)

is defined by

x_n=7-3n

for

n\ge1

. Write down its first three terms and state whether it is increasing or decreasing.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$4,1,-2$ Decreasing	2
Notes
Substitute $n=1,2,3$ to obtain $4,1,-2$ . Also $x_{n+1}-x_n=-3<0$ , so the sequence is decreasing.

(2 marks)

Tier 2 · Standard

The sequence is defined by

x_n=\dfrac{n}{n+1}

for

n\geq1

. Prove that the sequence is increasing.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$x_{n+1}-x_n=\dfrac{1}{(n+1)(n+2)}>0$ , so $(x_n)$ is increasing.	4
Notes
Calculate the difference between consecutive terms: $x_{n+1}-x_n=\dfrac{n+1}{n+2}-\dfrac{n}{n+1}=\dfrac{(n+1)^2-n(n+2)}{(n+1)(n+2)}=\dfrac{1}{(n+1)(n+2)}$ . For every $n\geq1$ the denominator is positive, so $x_{n+1}>x_n$ and the sequence is increasing.

(4 marks)

Tier 3 · Hard

A sequence is defined by

x_1=2

and

x_{n+1}=(x_n+6)/2

. Let

y_n=6-x_n

. Show that

(y_n)

is geometric, find a formula for

x_n

, and hence show that

(x_n)

is increasing.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$y_n=4\left(\dfrac{1}{2}\right)^{n-1}$ $x_n=6-4\left(\dfrac{1}{2}\right)^{n-1}$ $(x_n)$ is increasing	5
Notes
Using the recurrence, $y_{n+1}=6-x_{n+1}=6-(x_n+6)/2=(6-x_n)/2=y_n/2$ . Also $y_1=6-2=4$ , so $(y_n)$ is geometric with ratio $1/2$ and $y_n=4(1/2)^{n-1}$ . Therefore $x_n=6-4(1/2)^{n-1}$ . Since $(1/2)^{n-1}$ is positive and decreases as $n$ increases, subtracting it from $6$ shows that $(x_n)$ is increasing.

(5 marks)

Tier 1 · Easy

A sequence is defined by

x_1=14

and

x_{n+1}=\dfrac12x_n-1

. Find

x_2

x_3

and

x_4

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$x_2=6$ , $x_3=2$ , $x_4=0$	2
Notes
Apply the recurrence to the latest term each time: $x_2=14/2-1=6$ , $x_3=6/2-1=2$ , and $x_4=2/2-1=0$ .

(2 marks)

Tier 2 · Standard

For

n\geq1

, let

x_n=n^2-kn

, where

k

is real. Determine all values of

k

that make

(x_n)

increasing at every positive integer

n

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$k<3$	4
Notes
$x_{n+1}-x_n=(n+1)^2-k(n+1)-(n^2-kn)=2n+1-k$ . This expression is smallest when $n=1$ . It is positive for every $n\geq1$ precisely when $3-k>0$ , so $k<3$ .

(4 marks)

Tier 3 · Hard

For each positive integer

n

, let

x_n=n+\dfrac{36}{n}

. Prove that the terms decrease as far as

x_6

and increase thereafter. Hence identify the least term.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	The least term is $x_6=12$	5
Notes
$x_{n+1}-x_n=1+36/(n+1)-36/n=1-36/[n(n+1)]$ . This is negative when $n(n+1)<36$ , which holds for $n=1,\ldots,5$ , so $x_1>x_2>\cdots>x_6$ . It is positive for every $n\geq6$ because then $n(n+1)\geq42>36$ , so the sequence increases after $x_6$ . Therefore the least term is $x_6=6+36/6=12$ .

(5 marks)

Tier 2 · Standard

A sequence is defined by

x_1=k

and

x_{n+1}=\dfrac{x_n^2+12}{7}

, where

k

is real. Find every value of

k

for which the sequence is constant.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$k=3$ or $k=4$	4
Notes
For the sequence to be constant from its first term, $x_2=x_1=k$ . Hence $(k^2+12)/7=k$ , so $k^2-7k+12=0=(k-3)(k-4)$ . Thus $k=3$ or $k=4$ . Substituting either value into the recurrence returns the same value, so each produces a constant sequence.

(4 marks)

Tier 3 · Hard

A sequence is defined by

x_1=6

and

x_{n+1}=3+\dfrac{2}{x_n}

. Find the exact values of

x_2,x_3,\ldots,x_6

. Find both fixed points of the recurrence. By comparing only the terms you have calculated with the positive fixed point, describe their behaviour.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$x_2=\dfrac{10}{3}$ , $x_3=\dfrac{18}{5}$ , $x_4=\dfrac{32}{9}$ , $x_5=\dfrac{57}{16}$ , $x_6=\dfrac{203}{57}$ Fixed points $\dfrac{3+\sqrt{17}}{2}$ and $\dfrac{3-\sqrt{17}}{2}$ The calculated terms alternate on either side of the positive fixed point and move closer to it.	5
Notes
Repeated substitution gives $x_2=10/3$ , $x_3=18/5$ , $x_4=32/9$ , $x_5=57/16$ and $x_6=203/57$ . A fixed point $L$ satisfies $L=3+2/L$ , so $L^2-3L-2=0$ and $L=(3\pm\sqrt{17})/2$ . The positive fixed point is approximately $3.56155$ . The listed terms are approximately $6,3.33333,3.60000,3.55556,3.56250,3.56140$ , so they lie alternately above and below that value and their observed distances from it decrease.

(5 marks)

Tier 3 · Hard

A sequence is defined by

x_1=2

and

x_{n+1}=\dfrac{1}{1-x_n}

. Calculate

x_2,x_3,x_4

. Show algebraically that applying the recurrence three times returns any admissible term to its original value, and state the least period of this sequence. Hence find

x_{2026}

and the exact value of

\displaystyle\sum_{r=1}^{2025}x_r

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$x_2=-1$ , $x_3=\dfrac12$ , $x_4=2$ Least period $3$ $x_{2026}=2$ $\displaystyle\sum_{r=1}^{2025}x_r=\dfrac{2025}{2}$	7
Notes
Direct substitution gives $x_2=-1$ , $x_3=1/2$ and $x_4=2$ . If $f(x)=1/(1-x)$ , then $f(f(x))=(x-1)/x$ and $f(f(f(x)))=x$ whenever the expressions are defined. Since the three values $2,-1,1/2$ are distinct, the least period is $3$ . Since $2026$ leaves remainder $1$ when divided by $3$ , $x_{2026}=x_1=2$ . Each complete block sums to $2-1+1/2=3/2$ , and $2025=675(3)$ , so the required sum is $675(3/2)=2025/2$ .

(7 marks)

Q10

Tier 3 · Hard

10.

For positive integers

n

, a sequence is defined by

x_n=2\cos\left(\dfrac{n\pi}{3}\right)+(-1)^n

. Write down

x_1,x_2,\ldots,x_6

. Determine the least period of the sequence, justifying why no smaller positive period works. State all values taken by the sequence and describe exactly which positive integers

n

give each value. Hence decide whether the sequence is increasing, decreasing or neither.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$0,0,-3,0,0,3$ Least period $6$ $x_n=-3$ when $n$ leaves remainder $3$ on division by $6$ ; $x_n=3$ when it is divisible by $6$ ; otherwise $x_n=0$ Neither increasing nor decreasing	5
Notes
Substitution gives the first six terms $0,0,-3,0,0,3$ . Both $2\cos(n\pi/3)$ and $(-1)^n$ repeat after $6$ steps, so $6$ is a period. Period $1$ fails since $x_2=0$ but $x_3=-3$ ; period $2$ fails since $x_1=0$ but $x_3=-3$ ; period $3$ fails since $x_3=-3$ but $x_6=3$ ; period $4$ fails since $x_2=0$ but $x_6=3$ ; and period $5$ fails since $x_1=0$ but $x_6=3$ . Therefore the least period is $6$ . Each repeated block shows that the value is $-3$ at positions leaving remainder $3$ on division by $6$ , $3$ at positions divisible by $6$ , and $0$ otherwise. Since, for example, $x_2>x_3$ but $x_3<x_4$ , the sequence is neither increasing nor decreasing.

(5 marks)

Verified exam appearances

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Other points in 4 Sequences and series

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