Introduction
• This Chapter focuses on sequences and
series
• We will look at writing and using
algebraic sequences
• We will also be learning how to calculate
the sum of a sequence
Sequences and Series
The nth term
The nth term of a sequence is
sometimes known as the ‘general
term’.
You need to become familiar with
the terminology of sequences in Alevel maths.
U n  the nth term
U 3  the 3rd number in the sequence
Example 1
The nth term of a sequence is
given by Un = 3n – 1. Work out
the 1st, 3rd and 19th terms.
1st
3rd
U n  3n  1
U n  3n  1
U1  3  1
U3  9 1
U1  2
U3  8
19th
U n  3n  1
U 3  57  1
U 3  56
6B
Sequences and Series
The nth term
The nth term of a sequence is
sometimes known as the ‘general
term’.
You need to become familiar with
the terminology of sequences in Alevel maths.
U n  the nth term
U 3  the 3rd number in the sequence
Example 3
The nth term of a sequence is
given by:
n2
Un 
n 1
Work out the 20th term.
n2
Un 
n 1
20 2
U 20 
20  1
400
U 20 
21
6B
Sequences and Series
The nth term
The nth term of a sequence is
sometimes known as the ‘general
term’.
You need to become familiar with
the terminology of sequences in Alevel maths.
Example 3
Find the value of n for which
the formula
U n  5n  2
has a value of 153.
U n  5n  2
153  5n  2
U n  the nth term
155  5n
U 3  the 3rd number in the sequence
31  n
Un = 153
Add 2
Divide
by 5
6B
Sequences and Series
The nth term
The nth term of a sequence is
sometimes known as the ‘general
term’.
You need to become familiar with
the terminology of sequences in Alevel maths.
Example 4
Find the value of n for which
the formula
U n  n 2  7n  12
has a value of 72.
U n  n 2  7n  12
72  n  7 n  12
Un = 72
2
U n  the nth term
0  n 2  7 n  60
U 3  the 3rd number in the sequence
0  ( n  12)( n  5)
n  12 or n  5
But n has to be
positive, so n = 12
Subtract
72
Factorise
2 possible
solutions
6B
Sequences and Series
The nth term
Example 5
A sequence is generated by
the formula
The nth term of a sequence
is sometimes known as the
‘general term’.
You need to become familiar
with the terminology of
sequences in A-level maths.
1) Form 2 equations using the
information you have been
given
2) Solve them simultaneously
to find values for a and b
U n  an  b
Given that U3 = 5 and U8 = 20,
find the values of a and b.
n=3
U3 = 5
U n  an  b
U 3  3a  b
5  3a  b
5  3a  b
2) 20  8a  b
2) – 1) 15  5a
3a
4  b
n=8
U8 = 20
U n  an  b
U 8  8a  b
20  8a  b
1)
6B
Sequences and Series
Recurrence Relationships
When you have a rule to get from
one term to the next, you can use
a ‘recurrence relationship’
It is important to remember that
the sequences:
5, 8, 11, 14, 17
and
The rule could be described as
‘add 3 to the previous term’
4, 7, 10, 13, 16
U 2  U1  3
Will have the same recurrence
relationship:
The next
term
U3  U2  3
U4  U3  3
U k 1  U k  3
The current
term
5, 8, 11, 14, 17
U k 1  U k  3
However, the first one has U1 = 5
and the second has U1 = 4
6C
Sequences and Series
Recurrence Relationships
When you have a rule to get from
one term to the next, you can use
a ‘recurrence relationship’
Example 1
Find the first 5 terms of the
following sequences:
a) U n 1  U n  4
5, 8, 11, 14, 17
The rule could be described as
‘add 3 to the previous term’
U 2  U1  3
The next
term
U3  U2  3
U4  U3  3
U k 1  U k  3
The current
term
U1  7
7, 11, 15, 19, 23
b) U n 1  U n  4
U1  5
5, 9, 13, 17, 21
c) U n  2  3U n 1  U n U1  4 U 2  2
Next
term
Current Previous
term
term
4, 2, 2, 4, 10,
6C
Sequences and Series
Recurrence Relationships
When you have a rule to get from
one term to the next, you can use
a ‘recurrence relationship’
5, 8, 11, 14, 17
The rule could be described as
‘add 3 to the previous term’
U 2  U1  3
The next
term
U3  U2  3
U4  U3  3
U k 1  U k  3
The current
term
Example 2
A sequence of terms has the
following recurrence relationship:
U n  2  mU n 1  U n
U1  2 U 2  5
a) Find an expression for U3 in terms of m.
U n  2  mU n 1  U n
U 3  mU 2  U1
U 3  5m  2
Substitute n = 1
Put in the values
for U2 and U1
b) Find an expression for U4 in terms of m.
U n  2  mU n 1  U n
U 4  mU 3  U 2
U 4  m(5m  2)  5
U 4  5m  2m  5
2
Substitute n = 2
Put in the values
for U3 and U2
Simplify
6C
Sequences and Series
Arithmetic Sequences
A sequence that increases by a
constant amount is known as an
arithmetic sequence.
Example 1
Find the 10th, 50th and nth terms of
the following arithmetic sequence…
3, 7, 11, 15, 19…
First term: 3
3, 7, 11, 15, 19…(+4)
17, 14, 11, 8…(-3)
a, a + d, a + 2d, a + 3d…(+d)
Second term: 3 + 4
Third term: 3 + 4 + 4
Fourth term: 3 + 4 + 4 + 4
a) 10th term
b) 50th term
3 + (9 x 4)
3 + (49 x 4)
= 39
= 199
c) nth term
3 + ((n – 1) x 4)
= 3 + 4(n – 1)
6D
Sequences and Series
Arithmetic Sequences
A sequence that increases by a
constant amount is known as an
arithmetic sequence.
Example 2
Find the number of terms in the
following sequence.
7, 11, 15, …, …, 143
 Increases in 4s
3, 7, 11, 15, 19…(+4)
17, 14, 11, 8…(-3)
a, a + d, a + 2d, a + 3d…(+d)
 143 – 7 = 136
 136 ÷ 4 = 34
 So there are 34 ‘jumps’
7, 11, 15, …, …, 143
 There is always 1 more term
than there are jumps
 35 terms!
6D
Sequences and Series
The nth term of an arithmetic
sequence
All arithmetic sequences take the form:
a  ( a  d )  ( a  2d )  ( a  3d )  ( a  4 d ) etc...
Example 1
Find the 50th term of the following
sequences:
a) 4, 7, 10, 13…
a=4 d=3
and
b) 100, 93, 86, 79… a = 100
1st
term
2nd
term
3rd
term
4th
term
5th
term
We can put this together as a relationship
for the nth term of an arithmetic
sequence…
a  ( n  1) d
Where ‘a’ is the first term and ‘d’ is the
common difference.
d = -7
a) a  ( n  1) d
4  (50  1)  3
 151
b) a  ( n  1) d
100  (50  1)  ( 7)
 100  (49  7)
 243
6E
Sequences and Series
The nth term of an arithmetic
sequence
All arithmetic sequences take the form:
a  ( a  d )  ( a  2d )  ( a  3d )  ( a  4 d ) etc...
1st
term
2nd
term
3rd
term
4th
term
5th
term
We can put this together as a relationship
for the nth term of an arithmetic
sequence…
a  ( n  1) d
Where ‘a’ is the first term and ‘d’ is the
common difference.
Example 2
For the following sequence, calculate
the number of terms.
5 + 9 + 13 + 17 + 21 + … + 805
a=5
d=4
a  ( n  1) d  805
5  ( n  1)  4  805
5  4n  4  805
4n  1  805
4n  804
Substitute
numbers in
Work out the
bracket
Group together
terms
Subtract 1
Divide by 4
n  201
There are 201 terms in the sequence!
6E
Sequences and Series
The nth term of an arithmetic
sequence
All arithmetic sequences take the form:
a  ( a  d )  ( a  2d )  ( a  3d )  ( a  4 d ) etc...
1st
term
2nd
term
3rd
term
4th
term
5th
term
We can put this together as a relationship
for the nth term of an arithmetic
sequence…
a  ( n  1) d
Where ‘a’ is the first term and ‘d’ is the
common difference.
Example 3
Given that the 3rd term of an arithmetic
sequence is 20 and the 7th is 12:
a) Work out the first term
3rd term
7th term
a  ( n  1) d  20
a  (3  1) d  20
a  2d  20
a  ( n  1) d  12
a  (7  1) d  12
a  6d  12
a  2d  20
2) a  6d  12
1)
2) – 1) 4d  8
d  2
a  24
Substitute
into 1) or 2)
6E
Sequences and Series
The Sum of an Arithmetic Series
You need to be able to work out the sum
of numbers in an arithmetic sequence.
This method was
discovered by Carl
Friedrich Gauss (17771855) while he was still in
Primary School!
Add up the numbers from 1-100!
S  1  2  3  4  5  ...  99  100
S  100  99  ...  5  4  3  2  1
2S  101  101  101  ...  101  101
2 S  100 101
S  (100  101)  2
Write out the same
sequence backwards
Add both sequences
together
We have 100 lots of
101
Halve that to get the
actual total
S  5050
6F
Sequences and Series
The Sum of an Arithmetic Series
As a general rule:
S n  a  ( a  d )  ( a  2 d )  ( a  3d ) …, …, …  ( a  ( n  2) d )  ( a  ( n  1) d )
S n  ( a  ( n  1) d )  ( a  ( n  2) d ) …, …, …  ( a  3d )  ( a  2 d )  ( a  d )  a
a  ( a  ( n  1) d )
 2 a  ( n  1) d
a  d  ( a  ( n  2) d )
Group the a’s
 2 a  d  ( n  2) d
Multiply out
the bracket
 2a  d  nd  2d
 1) dthe d’s
2 a  ( n  1) d
2 S n  2 a  ( n  1) d  2 a  ( n  1) d …, …, …  2 a  ( nGroup
 2a  nd  d
Factorise
the
2nd part

2
a

(
n

1)
d
There
are
‘n
lots
of
2a
+
(n-1)d’
n
2
a

(
n

1)
d


2S 
n
n
 2a  (n  1)d 
Sn 
2
n
a  L
Sn 
2
Divide by 2
If L is the last term in the series
6F
Sequences and Series
The Sum of an Arithmetic Series
n
 2a  (n  1)d 
Sn 
2
Example 1
Calculate the value of the first 100 odd
numbers:
 1, 3, 5, 7, …, …
a=1
n
Sn 
a  L
2
a = the 1st term
d=2
n = 100
Sn 
n
 2a  (n  1)d 
2
Sn 
100
 2  (100  1)  2
2
d = the common
difference
S n  50  2  198
L = the last term
S n  50  200 
S n  10, 000
Substitute
numbers in
Work out the
inner bracket
50 x 200
6F
Sequences and Series
The Sum of an Arithmetic Series
n
 2a  (n  1)d 
Sn 
2
Example 1
Calculate the value of the first 100 odd
numbers:
 1, 3, 5, 7, …, …
a=1
n
Sn 
a  L
2
a = the 1st term
d = the common
difference
L = the last term
d=2
n = 100
100th term  a  ( n  1) d
1  (100  1)  2
 199
n
Sn 
a  L
Substitute
2
numbers in
100
Sn 
1  199 
2
S n  50  200 
S n  10, 000
6F
Sequences and Series
The Sum of an Arithmetic Series
n
 2a  (n  1)d 
Sn 
2
Example 2
Find the number of terms needed for the sum
of the following sequence to exceed 2000.
4 + 9 + 14 + 19…
Sn = 2000
n
Sn 
a  L
2
a=4
n
 2a  (n  1)d 
2
n
2000  8  ( n  1)  5
2
Sn 
a = the 1st term
4000  n 8  5n  5
d = the common
difference
4000  n  3  5n 
L = the last term
4000  5n 2  3n
0  5n  3n  4000
2
d=5
Substitute
numbers in
Multiply by 2
Group together
terms
Multiply out the
bracket
Subtract 4000
6F
Sequences and Series
The Sum of an Arithmetic Series
n
 2a  (n  1)d 
Sn 
2
Example 2
Find the number of terms needed for the sum
of the following sequence to exceed 2000.
4 + 9 + 14 + 19…
0  5n 2  3n  4000
n
Sn 
a  L
2
a = the 1st term
d = the common
difference
L = the last term
a=5
b  b 2  4ac
2a
3  9  (4  5  4000)
10
3  9  ( 80000)
10
3  80009
10
b=3
c = -4000
Substitute
numbers in
Careful with
negatives!
n = 27.9 or -28.5
n = 28 ( need 28 terms
to be over 2000)
6F
Sequences and Series
Sequences Notation
The symbol Σ can be used to mean
‘sum of’:
Highest value
of r
Highest value
of r
10
r 15
r 0
r 5
 (2  3r )
The first
value of r
The formula
to be used
 (10  2r )
The first
value of r
The formula
to be used
So this means the sum of the
sequence (2 + 3r) from r = 0 to r = 10
So this means the sum of the
sequence (10 - 2r) from r = 5 to r = 15
 Sum of 2 + 5 + 8 + … + 32
 Sum of 0 + -2 + -4 + … + -20
6G
Sequences and Series
Sequences Notation
The symbol Σ can be used to mean
‘sum of’:
Example 1
Calculate the value of the following:
r  20
 4r  1
Highest value
of r
r 1
5 + 9 + 13 + … + 81
10
 (2  3r )
r 0
The first
value of r
The formula
to be used
Remember that the formulae only work for ARITHMETIC
sequences!
 Sometimes on exams you have to work out the sum of
the first 4-5 terms, but the sequence is NOT arithmetic
 In this case, just work out each term and add them up
manually!
Sn 
n
 2a  (n  1)d 
2
Sn 
20
10  (20  1)  4 
2
S n  10 10  19  4 
a=5
d=4
n = 20
Sub
numbers
in
Work out
brackets
S n  10 86 
S n  860
6G
Summary
• We have looked at sequences
• We have seen how to calculate a number in an
arithmetic sequence
• We have also worked out the sum of a
sequence
• We have also seen some of the notation which
is used in sequences