This document discusses different types of sequences in mathematics, including linear, quadratic, and geometric sequences. It provides examples and explanations of how to determine the rule that generates each sequence, either from term-to-term or from the position of a term in the sequence. Key points covered include finding the common difference or second difference to classify a sequence as linear or quadratic, and determining the formula for the nth term of a sequence from its pattern of terms. Examples are given of predicting future terms and finding specific terms of various sequences.

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Sequences
Mathematics

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AA7.1 Generating sequences from rules
A7 Sequences
Contents
A7.2 Linear sequences
A7.3 Quadratic sequences
A7.4 Geometric sequences
A7.5 Other types of sequence

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2, 5, 8, 11, 14, 17, 20, 23, . . .
In mathematics, a sequence is a succession of numbers that
follow a given rule.
Each number in a sequence is called a term.
If terms are next to each other they are referred to as
consecutive terms.
1st term 6th term
Sequences
For example,

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Predicting terms in a sequence
Sometimes, we can predict how a sequence will continue by
looking for patterns.
For example,
What are the next two terms in the following sequence,
102, 95, 88, 81, 74 . . . ?
We can predict that this sequence continues by subtracting
7 each time. We can use this to find the next two terms.
Look at the difference between each consecutive term.
–7 –7 –7 –7
102 95 88 81 74
–7
67
–7
60

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Predicting terms in a sequence
If we are not given the rule for a sequence, or if it is not
generated from a practical context, we cannot be certain how
it will continue.
For example, a sequence starts with the numbers 2, 4, 8, . . .
How could this sequence continue?
+2 +4
2 4 8
+6
14
+8
22
+10
32
+12
44
×2 ×2
2 4 8
×2
16
×2
32
×2
64
×2
128

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There are two ways to define a sequence.
Defining sequences
The first is to use a term-to-term rule.
To define a sequence using a term-to-term rule we need to
know the first term in the sequence and what must be done
to each term to give the value of the next term.
The second is to use a position-to-term rule.
To define a sequence using a position-to-term rule we use
a formula for the nth term of the sequence.
Term-to-term rules are usually easier to find for a given
sequence.
Position-to-term rules are harder to find for a given sequence
but are more useful for finding any term in a sequence.

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Write the first five terms of each sequence
given the first term and the term-to-term rule.
1st term Term-to-term rule
5
68
48
400
–1.1
0.1776
Sequences from term-to-term rules
–7 Add 3
100 Subtract 8
3 Double
0.04 Multiply by 10
1.3 Subtract 0.6
111 Divide by 5
–7, –4, –1, 2,
100, 92, 84, 76,
3, 6, 12, 24,
0.04, 0.4, 4, 40,
1.3, 0.7, 0.1, –0.5,
111, 22.2, 4.44, 0.888,
First five terms

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When a sequence is defined by a position-to-term rule it can
sometimes help to put the terms in a table. For example,
Sequences from position-to-term rules
Each term can be found by squaring its position in the
sequence.
What is the 20th term in this sequence?
202 = 400
The nth term in a sequence is n2, where n is the term’s
position in the sequence.
Position
Term
1st 2nd 3rd 4th 5th …
…
nth
n2
1 4 9 16 25

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Writing sequences from position-to-term rules
The nth term, or the general term, of a sequence is often
called un.
the 2nd term u2,
the 3rd term u3,
the 4th term u4,
The 1st term is then called u1,
Any term in the sequence can then be found by substituting
its position number into the formula for un.
the 5th term u5 and so on.

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For example, suppose the nth term of a sequence is 4n – 5.
We can write this rule as:
un = 4n – 5
Find the first five terms in the sequence.
u1 = 4 × 1 – 5 = –1
u2 = 4 × 2 – 5 = 3
u3 = 4 × 3 – 5 = 7
u4 = 4 × 4 – 5 = 11
u5 = 4 × 5 – 5 = 15
The first five terms in the sequence are: –1, 3, 7, 11 and 15.
Writing sequences from position-to-term rules

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The nth term of a sequence is 2n2 + 3.
We can write this rule as:
un = 2n2 + 3
Find the first 4 terms in the sequence.
u1 = 2 × 12 + 3 = 5
u2 = 2 × 22 + 3 = 11
u3 = 2 × 32 + 3 = 21
u4 = 2 × 42 + 3 = 35
The first 4 terms in the sequence are: 5, 11, 21, and 35.
Writing sequences from position-to-term rules
This sequence is a quadratic sequence.

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Writing sequences from position-to-term rules

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Matching sequences to rules

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A7.2 Linear sequences
Contents
A7 Sequences
A7.3 Quadratic sequences
A7.4 Geometric sequences
A7.1 Generating sequences from rules
A7.5 Other types of sequence

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Sequences that increase in equal steps
To work out a rule for a sequence it is often helpful to find the
difference between consecutive terms.
For example, look at the difference between each term in
this sequence:
2, 8, 14, 20, 26, 32, 38, 44, . . .
+6 +6 +6 +6 +6 +6 +6
This sequence starts with 2 and increases by adding 6 each
time.
Every term in this sequence is 4 less than a multiple of 6.

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Can you work out the next three terms in this sequence?
How did you work these out?
This sequence starts with 10 and decreases by subtracting 3
from each term to give the next term.
Sequences that decrease in equal steps
10, 7, 4, 1, –2, –5, –8, –11, . . .
–3 –3 –3 –3 –3 –3 –3
When the difference between each term in a sequence is
always the same we have a linear or arithmetic sequence.

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When we describe linear sequences we call the constant
difference between consecutive terms the common
difference, d.
We call the first term in a linear sequence a.
If we are given the values of a and d in a linear sequence
then we can use them to generate the sequence.
For example, if a linear sequence has a = 5 and d = –3,
we have the sequence:
5, 2, –1, –4, –7, –10, . . .
Generating linear sequences
Linear sequences can also be generated given the rule for
the nth term in the sequence.

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Suppose we are given a linear sequence and asked to
find the nth term, un, of the sequence. For example,
Find the nth term of the sequence 4, 7, 10, 13, 16, …
The nth term of a linear sequence
This sequence continues by adding 3 each time and so
the common difference d is 3.
We compare the terms in the sequence to the multiples of 3.
Position
Multiples
of 3
1 2 3 4 5 n
…
× 3 × 3 × 3 × 3 × 3 × 3
3n
Term 4 7 10 13 16 …
3 6 9 12 15
+ 1 + 1 + 1 + 1 + 1 + 1
3n + 1
un = 3n + 1.

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Find the nth term of the sequence 5, 3, 1, –1, –3, …
The nth term of a linear sequence
This sequence continues by subtracting 2 each time and
so the common difference d is –2.
We compare the terms in the sequence to the multiples of –2.
Position
Multiples
of –2
1 2 3 4 5 n
…
× –2 × –2 × –2 × –2 × –2 × –2
–2n
Term 5 3 1 –1 –3 …
–2 –4 –6 –8 –10
+ 7 + 7 + 7 + 7 + 7 + 7
7 – 2n
un = 7 – 2n.

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If a is the first term of a linear sequence and d is the common
difference, we can find the general form of the nth term as
follows:
The nth term of a linear sequence
u1 = a,
u2 = a + d,
u3 = a + 2d,
u4 = a + 3d,
un = a + (n – 1)d,
Multiplying out the bracket we have un = a + dn – d.
The nth term of a linear sequence with first term a and common
difference d is:
un = dn + (a – d)
(a – d) is the value
of the 0th term.

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Position number (n)
Value
of
term
(u
n
)
We can also find the nth term of a linear sequence by plotting
the value of each given term in the sequence against its
position number. For example,
Plotting terms
The points lie on a straight line.
The equation of the line can be
written in the form y = mx + c.
un = 2n + 1
The sequence 3, 5, 7, 9, 11 … can be shown graphically:
The value of the gradient m corresponds
to the difference between the terms.
The value of the y-intercept c
corresponds to the 0th term, to give

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Finding the nth term of a linear sequence

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Tiling patterns
The following patterns are made from tiles:
How many tiles will there be in the next pattern?
How many tiles will there be in the 20th pattern?
Pattern 1
1 tile Pattern 2
5 tiles Pattern 3
9 tiles Pattern 4
13 tiles

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Tiling patterns
To work out the number of tiles in the next pattern we can
look at the difference between consecutive terms in the
sequence:
1, 5, 9, 13,
+4 +4 +4 +4
17
The difference between each term is always equal to 4 and
so we can use this to find the next term.
Using this pattern, we can predict that there will be 17 tiles in
the next pattern.

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Tiling patterns
To work out how many tiles there will be in the 20th pattern
we can find the rule for the nth term in the sequence.
1, 5, 9, 13,
+4 +4 +4 +4
17
This is a linear sequence and so the nth term will be of the
form un = dn + c, where d is the constant difference between
the terms.
Look at the pattern made by the differences:
How can we find the value of c?
For this sequence the constant difference d = 4.

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Tiling patterns
We can find the rule for the nth term by comparing the
sequence to multiples of 4.
Multiples
of 4
Position 1 2 3 4 5 n
…
× 4 × 4 × 4 × 4 × 4 × 4
4n
Term 1 5 9 13 17 …
4 8 12 16 20
– 3 – 3 – 3 – 3 – 3 – 3
4n – 3
un = 4n – 3.
We can use this formula to predict the number of tiles in the
20th pattern:
u20 = 4 × 20 – 3 = 77

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A7.3 Quadratic sequences
Contents
A7 Sequences
A7.4 Geometric sequences
A7.2 Linear sequences
A7.1 Generating sequences from rules
A7.5 Other types of sequence

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Some sequences increase or decrease in unequal steps.
Sequences that increase in increasing steps
For example, look at the differences between terms in this
sequence:
4, 5, 7, 10, 14, 19, 25, 32, . . .
+1 +2 +3 +4 +5 +6 +7
This sequence starts with 4 and increases by adding
consecutive whole numbers to each term.
The differences between the terms form a linear sequence.

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Can you work out the next three terms in this sequence?
How did you work these out?
This sequence starts with 7 and decreases by subtracting
0.1, 0.2, 0.3, 0.4, 0.5, …
Sequences that decrease in decreasing steps
7, 6.9, 6.7, 6.4, 6, 5.5, 4.9, 4.2, . . .
–0.1 –0.2 –0.3 –0.4 –0.5 –0.6 –0.7
With sequences of this type it is often helpful to find a second
row of differences.

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Can you work out the next three terms in this sequence?
Look at the differences between terms.
Using a second row of differences
1, 3, 8, 16, 27, 41, 58, 78
+2 +5 +8 +11 +14 +17 +20
A sequence is formed by the first row of differences so we
look at the second row of differences.
+3 +3 +3 +3 +3 +3
This shows that the differences increase by 3 each time.
First row of
differences
Second row of
differences

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Quadratic sequences
When the second row of differences produces a constant
number, the sequence is called a quadratic sequence.
This is because the rule for the nth term of the sequence is a
quadratic expression of the form
un = an2 + bn + c
where a, b and c are constants and a ≠ 0.
The simplest quadratic sequences is the sequence of square
numbers.
The constant second difference is 2 and the nth term is n2.
1, 4, 9, 16, 25
+3 +5 +7 +9
+2 +2 +2

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Investigating quadratic sequences

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When un = an2 + bn + c the value of a can be
found by halving the value of the second
difference, which is always equal to 2a.
The nth term of a quadratic sequence
We can prove this as follows,
un = an2 + bn + c
u1 = a × 12 + b × 1 + c =
u2 = a × 22 + b × 2 + c =
u3 = a × 32 + b × 3 + c =
u4 = a × 42 + b × 4 + c =
This gives us the first five terms of a general quadratic sequence.
u5 = a × 52 + b × 5 + c =
a + b + c
4a + 2b + c
9a + 3b + c
16a + 4b + c
25a + 5b + c

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The nth term of a quadratic sequence
Let’s find the first and second differences for these general
terms.
a + b + c 4a + 2b + c 9a + 3b + c 16a + 4b + c 25a + 5b + c
3a + b 5a + b 7a + b 9a + b
2a 2a 2a
This shows that the second difference is always 2a when
un = an2 + bn + c.

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When un = an2 + bn + c the value of c can be found by
calculating the value of the 0th term, u0.
The nth term of a quadratic sequence
We can prove this as follows,
un = an2 + bn + c
u0 = a × 02 + b × 0 + c
Of course, the 0th term is not really part of the sequence. We
can easily work it out though by looking at the sequence
formed by the first row of differences and counting back from
the first term.
u0 = c

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When un = an2 + bn + c the value of b can be found by
subtracting (a + c) from the value of the first term.
The nth term of a quadratic sequence
We can prove this as follows,
un = an2 + bn + c
u1 = a × 12 + b × 1 + c
Rearranging this formula to make b the subject gives,
u1 = a + b + c
b = u1 – (a + c)
If we find the value of a and c first, we can then use them and
the value of the first term, to find b.

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The nth term of a quadratic sequence
Find the nth term of the sequence, 4, 9, 18, 31, 48, …
Let’s start by looking at the first and second differences.
4, 9, 18, 31, 48
+5 +9 +13 +17
+4 +4 +4
The second differences are constant and so the nth term is in
the form un = an2 + bn + c.
Let’s find a, b and c.
The second difference is 4, so we know 2a = 4
a = 2

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The nth term of a quadratic sequence
Let’s start by looking at the first and second differences.
4, 9, 18, 31, 48
+5 +9 +13 +17
+4 +4 +4
The value of c is the same as the value for the 0th term.
+4
+1
3,
We can find this by continuing the pattern in the differences
backwards from the first term.
The 0th term is 3, so:
c = 3
Find the nth term of the sequence, 4, 9, 18, 31, 48, …

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The nth term of a quadratic sequence
Putting a = 2 and c = 3 into un =an2 + bn + c gives us
un = 2n2 + bn + 3. We can use this to write an expression for
the first term:
u1 = 2 + b + 3
u1 = b + 5
The first term in the sequence is 4, so:
4 = b + 5
u1 = 2 × 12 + b × 1 + 3
un = 2n2 + bn + 3
b = –1
Find the nth term of the sequence, 4, 9, 18, 31, 48, …

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The nth term of a quadratic sequence
Once we have found the values of a, b, and c we can use
them in un = an2 + bn + c to give the nth term.
We have found that for the sequence 4, 9, 18, 31, 48, …
a = 2, b = –1 and c = 3,
un = 2n2 – n + 3
We can check this rule by substituting a chosen value for n
into the formula and making sure that it corresponds to the
required term in the sequence.
For example, when n = 5 we have,
u5 = 2 × 52 – 5 + 3
= 48
Check the rule for other terms in the sequence.

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The nth term of a quadratic sequence
Find the nth term of the sequence 1, 3, 6, 10, 15, …
This is the sequence of triangular numbers.
1, 3, 6, 10, 15
+2 +3 +4 +5
+1 +1 +1
The second differences are constant and so the nth term is in
the form un =an2 + bn + c.
Let’s find a, b and c.
The second difference is 1, so 2a = 1
a = ½

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The nth term of a quadratic sequence
Find the nth term of the sequence 1, 3, 6, 10, 15, …
This is the sequence of triangular numbers.
1, 3, 6, 10, 15
+2 +3 +4 +5
+1 +1 +1
The value of c is the same as the value for the 0th term.
+1
+1
0,
We can find this by continuing the pattern in the differences
backwards from the first term.
The 0th term is 0, so:
c = 0

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The nth term of a quadratic sequence
Find the nth term of the sequence 1, 3, 6, 10, 15, …
Putting a = ½ and c = 0 into un =an2 + bn + c gives us
un = ½n2 + bn. We can use this to write an expression for the
first term:
u1 = ½ + b
The first term in the sequence is 1, so
1 = ½ + b
u1 = ½ × 12 + b × 1
un = ½n2 + bn
b = ½

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The nth term of a quadratic sequence
Once we have found the values of a, b, and c we can use
them in an2 + bn + c to give the nth term.
We have found that for the sequence 1, 3, 6, 10, 15, …
a = ½, b = ½ and c = 0;
Checking, when n = 5, we have
u5 = ½(52 + 5)
= 15
Check the rule for other terms in the sequence.
un =
n2
2
+
n
2
or
un =
n2 + n
2

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Finding the nth term of a quadratic sequence

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Tiling patterns
The following patterns are made from tiles,
Pattern 1
1 tile Pattern 2
5 tiles Pattern 3
13 tiles Pattern 4
25 tiles
How many tiles will there be in the next pattern?
How many tiles will there be in the 20th pattern?

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Tiling patterns
To work out the number of tiles in the next pattern we can
look at the difference between consecutive terms in the
sequence,
1, 5, 13, 25,
+4 +8 +12 +16
+4 +4 +4
41
If the second difference is always equal to 4 then we can use
this to find the next term.
Using this pattern we can predict that there will be 41 tiles in
the next pattern.

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Tiling patterns
To work out how many tiles there will be in the 20th pattern
we can find the rule for the nth term in the sequence.
The second difference is constant and so the nth term will be
a quadratic of the form un = an2 + bn + c.
The second difference is always equal to 2a:
1, 5, 13, 25,
+4 +8 +12 +16
+4 +4 +4
41
Look at the pattern made by the differences,
2a = 4
a = 2

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Tiling patterns
To work out how many tiles there will be in the 20th pattern
we can find the rule for the nth term in the sequence.
1, 5, 13, 25,
+4 +8 +12 +16
+4 +4 +4
41
The value of c is equal to the 0th term.
Look at the pattern made by the differences,
The 0th term is equal to 1, so:
c = 1
+4
+0
1,

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Tiling patterns
To work out how many tiles there will be in the 20th pattern
we can find the rule for the nth term in the sequence.
1, 5, 13, 25,
+4 +8 +12 +16
+4 +4 +4
41
If a = 2 and c = 1 then un = 2n2 + bn + 1,
Look at the pattern made by the differences,
u1 = 2 + b + 1
u1 = b + 3
1 = b + 3
b = –2
1 is the first
term.

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Tiling patterns
We can now substitute a = 2
b = –2
c = 1
into the general form for the nth term of the quadratic sequence,
an2 + bn + c, to give us the following rule for the nth term:
We can now use this formula to find the value of the 20th
term.
un = 2n2 – 2n + 1
u20 = 2 × 202 – 2 × 20 + 1
= 800 – 40 + 1
= 761
We can predict that there will be 761 tiles in the 20th pattern.

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A7.4 Geometric sequences
Contents
A7 Sequences
A7.3 Quadratic sequences
A7.2 Linear sequences
A7.1 Generating sequences from rules
A7.5 Other types of sequence

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Some sequences increase or decrease by multiplying or
dividing each term by a constant factor.
Sequences that increase by multiplying
For example, look at this sequence:
2, 4, 8, 16, 32, 64, 128, 256, . . .
×2 ×2 ×2 ×2 ×2 ×2 ×2
This sequence starts with 2 and increases by multiplying the
previous term by 2.
All of the terms in this sequence are powers of 2.

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Can you work out the next three terms in this sequence?
How did you work these out?
This sequence starts with 1024 and decreases by dividing
by 4 each time.
Sequences that decrease by dividing
1024, 256, 64, 16, 4, 1,
÷4 ÷4 ÷4 ÷4 ÷4 ÷4 ÷4
We could also continue this sequence by multiplying each
term by to give the next term.
1
4
, . . .
1
16
1
4
,

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Geometric sequences
Sequences that go from one term to the next by multiplying
by a constant amount are called geometric sequences.
The amount that each term is multiplied by to get the next
term is called the common ratio of the sequence.
The common ratio r of a geometric sequence can be found
by dividing any term in the sequence by the one before it.
For example,
Find the common ratio for the following geometric sequence:
8, 12, 18, 27, 40.5, …
r = 12 ÷ 8 = 1.5
The sequence continues by multiplying the previous term by 1.5.

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Continuing geometric sequences

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The nth term of a geometric sequence
Suppose we are given a geometric sequence and asked
to find the nth term un of the sequence. For example:
Find the nth term of the sequence, 3, 6, 12, 24, 48, …
The first term of this sequence is 3 and it continues by
multiplying the previous term by 2 each time. We can write
this sequence as follows:
u1 = 3
u2 = 6 = 3 × 2
u3 = 12 = 3 × 2 × 2 = 3 × 22
u4 = 24 = 3 × 2 × 2 × 2 = 3 × 23
u5 = 24 = 3 × 2 × 2 × 2 × 2 = 3 × 24
un = 3 × 2(n – 1)

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The nth term of a geometric sequence
In general, if the first term of a geometric sequence is a and
the common ratio is r, then we can find the general form of
the nth term as follows:
u1 = a,
u2 = a × r,
u3 = a × r2,
u4 = a × r3,
un = a × rn – 1,
The nth term of a geometric sequence with first term a and
common ratio r is
un = arn – 1
u5 = a × r4,

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Geometric sequences in real-life
Geometric sequences often occur in real-life situations
where there is a repeated percentage change.
Write a formula for the value of the
investment at the beginning of the nth year.
For example, £800 is invested in an account with an annual
interest rate of 5%.
Every year the amount in the account is multiplied by 1.05.
The amount in the account at the beginning of each year
forms a geometric sequence with £800 as the first term and
1.05 as the common ratio.
£800,
×1.05 ×1.05 ×1.05
£840, £882, £926.10, …

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Geometric sequences in real-life
Geometric sequences often occur in real-life situations
where there is a repeated percentage change.
Write a formula for the value of the
investment at the beginning of the nth year.
For example, £800 is invested in an account with an annual
interest rate of 5%.
We can write the sequence as,
u1 = £800
u2 = £800 × 1.05
u3 = £800 × 1.052
u4 = £800 × 1.053
un = £800 × 1.05n – 1

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Contents
A7.5 Other types of sequence
A7 Sequences
A7.4 Geometric sequences
A7.3 Quadratic sequences
A7.2 Linear sequences
A7.1 Generating sequences from rules

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Fraction sequences
The terms in the following sequence are all fractions:
2
3
, 3
6
, 4
9
, 5
12
, 6
15
, …
Find the next two terms in the sequence.
Look at the sequence formed by the numerators and the
sequence formed by the denominators separately.
3
2 ,
6
3 ,
9
4 ,
12
5 ,
15
6 ,
+3 +3 +3 +3 +3 +3
+1 +1 +1 +1 +1 +1
18
7 ,
21
8 , …

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Fraction sequences
The terms in the following sequence are all fractions,
2
3
, 3
6
, 4
9
, 5
12
, 6
15
, …
Find the formula for the nth term of the sequence.
The sequence formed
by the numerators is:
u1 = 2
u2 = 3
u3 = 4
un = n + 1
The sequence formed
by the numerators is:
u1 = 3
u2 = 6
u3 = 9
un = 3n

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Fraction sequences
The terms in the following sequence are all fractions:
2
3
, 3
6
, 4
9
, 5
12
, 6
15
, …
Find the formula for the nth term of the sequence.
Given that the nth term for the numerators is n + 1 and the nth
term for the denominators is 3n, we can write the formula for
the nth term of the above sequence as:
un =
n + 1
3n

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The Fibonacci Sequence
Can you work out the next three terms in this sequence?
How did you work these out?
This sequence starts 1, 1 and each term is found by
adding together the two previous terms.
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
This sequence is called the Fibonacci Sequence after the
Italian mathematician who first wrote about it.
1+1 1+2 2+3 3+5 5+8 8+13 13+21 21+34

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The nth term of the Fibonacci Sequence
The Fibonacci Sequence is an example of a sequence for
which the general term cannot be written in terms of its
position in the sequence.
To find any given term we have to know the value of
the previous two terms.
We can write: u1 = 1
u2 = 1
u3 = u2 + u1
u4 = u3 + u2
u5 = u4 + u3
The nth term can then be written as:
un = un – 1 + un – 2

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The nth term of the Fibonacci sequence
The formula is called an iterative formula.
Can you use this formula to find the
100th term in the sequence?
un = un – 1 + un – 2
The sequence cannot be defined by this formula alone. We
have to be given the value of the first two terms.
Using this formula would give us,
u100 = u99 + u98
All this tells us is that the 100th term is equal to the sum of the
99th term and the 98th term.
The only way to find the 100th term is to use a computer or
spend some time doing a lot of arithmetic!

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Using iterative formulae to generate sequences
Write down the first five terms of the
sequence generated by the iterative formula
un = 2un – 1 + 1
when u1 = 3.
Can you find the formula for the
nth term of this sequence?
u1 = 3
u2 = 6 + 1 = 7
u3 = 14 + 1 = 15
u4 = 30 + 1 = 31
u5 = 62 + 1 = 63
The first five terms of the sequence are 3, 7, 15, 31, 63, …