SEQUENCES AND ARITHMETIC SERIES
1. Sequences
A sequence is a set of numbers with either a formula for the nth term, or a rule for getting from
one term to the next. Consider the sequence 1, 3, 5, 7, 9, 11, ... (the dots indicate the sequence
goes on for ever). We use u1 to denote the first term, u2 to denote the second, un to denote the
nth term etc. So in this case u1 = 1, u2 = 3, u3 = 5 etc. We can define this sequence in two ways
a) Recurrence Relation
u1  1, un1  un  2
The recurrence relation tells us what the first term is, and gives us a rule for getting from one
term to the next. Using this definition the whole sequence can be built up. However, this
method has its limitations - to find the 263rd term ( u263 ), we would have to work out the
previous 262 terms, one by one, first.
b) Algebraic Definition
un  2 n  1
This definition tells us how to work out the nth term straight away. For example,
u263  2  263  1
 525
Some more sequences...
sequence
2, 7, 12, 17, 22...
recurrence relation
u1  2, un1  un  5
algebraic definition
un  5n  3
94, 81, 68, 55, 42...
u1  94, un1  un  13
un  107  13n
2, 4, 8, 16, 32...
u1  2, un1  2un
un  2 n
8, 16, 32, 64, 128...
u1  8, un1  2un
un  4  2n or un  2n  2
1, 4, 9, 16, 25...
u1  1, un 1 
1
4
,
1
5
,
1
6
,
1
7
,
1
8
…
1, –3, 9, –27, 81…

u1  14 , un 1 

un  1
2
un
un  1
u1  1, un1  3un
un  n 2
un 
1
n3
un  (3) n 1
C1 p83 Ex 6A, p87 Ex 6C
2. Behaviour of a Sequence
Sequences can show one of three types of behaviour...
a) Convergent
1
Consider the sequence un  2  . The sequence begins 3, 2½, 2⅓, 2¼, ... It is clear that the
n
terms get closer and closer to 2, without ever getting there (see fig. 1). We say the sequence
converges to 2, or that the limit of the sequence is 2. We use this notation...
1
lim 2   2
n 
n
Another convergent sequence is un 
n 1
. The sequence begins 0, 13 ,
n 1
2
4
, 53 ... In this case the
limit of the sequence is 1 (see fig. 2).
 
Some sequences have oscillatory convergence e.g. un   12
 18 ,
1
16
n
. The sequence begins  12 ,
1
4
,
,... In this case the limit of the sequence is 0 (see fig. 3).
3
2
u
u
1
0
1
0.8
0.6
0.4
0.2
u
0.4
0.2
0
0
-0.2 1 2 3 4 5 6 7 8 9 10
-0.4
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
n
n
n
fig.1
fig 2
fig.3
-0.6
b) Divergent
Such sequences have no limit, and their terms tend to infinity (either positive or negative).
An example is un  2n . The terms get larger and larger with no limit (see fig. 4).
Some sequences demonstrate oscillatory divergence e.g. un  (2) n . The sequence exhibits
larger and larger oscillations, and once again there is no limit (see fig. 5).
20
1500
15
1000
500
u 10
u
0
5
-500 1 2 3 4 5 6 7 8 9 10
0
-1000
1 2 3 4 5 6 7 8 9 10
n
n
fig. 4
fig. 5
c) Periodic
Such sequences neither converge to a limit, nor diverge to infinity, but repeat themselves every
so often. Consider the sequence un  ( 1) n . The sequence begins –1, 1, –1, 1, –1 ... We say
the sequence is periodic with period 2 i.e. it repeats itself every 2 terms (see fig. 6)
1
Another example is u1  2, un 1  1  . The sequence begins 2, ½, –1, 2, ½, –1... and
un
therefore is periodic with period 3 (see fig. 7).
2
3
1
2
1
u 0
-1
u
1 2 3 4 5 6 7 8 9 10
-2
0
-1 1 2 3 4 5 6 7 8 9 10
-2
n
n
fig. 6
fig.7
3. Using the Algebraic Definition
Example 1 : Find the number of terms in the sequence 3, 7, 11, 15, ... , 351.
The algebraic definition of the sequence is
un  4 n  1
For the last term,
351  4n  1
n  88
So there are 88 terms in the sequence.
Example 2 : Find the number of terms in the sequence 27, 21, 15, 9, ... , –51.
The algebraic definition of the sequence is
un  33  6n
For the last term,
33  6n  51
n  14
So there are 14 terms in the sequence.
Example 3 : At which term does the sequence 36, 43, 50, 57, ... first exceed 1000?
un  7n  29
7 n  29  1000
7n  971
n  138.714...
n  139
th
The sequence first exceeds 1000 at the 139 term, which is (7 × 139) + 29 = 1002.
Example 3 : The sequence with formula un  an  b has u3  10 and u7  18 . Find the values of
a and b.
Substituting values of n, we get two simultaneous equations.
3a  b  10
7a  b  18
Subtracting,
4a  8
a2
Substituting back,
3  2  b  10
b4
C1 p85 Ex 6B
4. Series and Sigma Notation
The sum of the terms of a sequence is known as a series.
So 1, 3, 5, 7, 9 is a sequence, and 1 + 3 + 5 + 7 + 9 is a series, whose sum is 25.
Sigma notation is a concise way of writing series. The series 1 + 3 + 5 + 7 + 9 becomes
5
 2r  1
r 1
in sigma notation (Σ is the Greek letter sigma). This means ‘add together all the numbers worth
2r  1 where r takes all integer values from 1 to 5 inclusive.’
The sigma notation representation is not unique. For example, 1 + 3 + 5 + 7 + 9 can also be
written as
4
 2r  1
r 0
11
or
 2r  13
r 7
However, it is more convenient to have r = 1 as the lower limit. Some more examples...
4
2  4  8  16   2r
r 1

1
r 1
r 1 2
Notice that in the last example, an infinite number of terms is being summed.
1
4
 18  161  321 ...  
5. Arithmetic Progressions (A.P.s)
An arithmetic progression is a sequence in which each term differs from the previous term by a
fixed amount, known as the common difference.
•
•
4, 7, 10, 13, 16, ... is an arithmetic progression with common difference 3.
16, 14, 12, 10, 8, ... is an arithmetic progression with common difference −2.
In general an A.P. with first term a and common difference d has terms
a, a + d, a + 2d, a + 3d, ......
In particular,
nth term of an arithmetic progression  a  (n  1)d
If an A.P. has a finite number of terms, we can find the sum of its terms. Consider the A.P. with
first term 3 and common difference 4. Suppose we wish to find the sum of the first ten terms S10 .
S10  3  7  11  15  19  23  27  31  35  39
S10  39  35  31  27  23  19  15  11  7  3
2S10  42  42  42  42  42  42  42  42  42  42
2S10  10  42
10  42
2
 210
S10 
By reversing the series and adding, we have ten terms, all of the same size.
Clearly this method will work with any A.P. Suppose we wish to find the sum of the first 37
terms of the A.P. with first term 51 and common difference –5.
u37  51  (36  5)
 129
S37  51  46  41  ....  129
S37  129  124  119  .............  51
2 S37  37  78
37  78
2
 1443
We can generalise, using this method to derive a formula for the sum of the first n terms of an
A.P. with first term a and common difference d.
S37 
Sn  a  (a  d )  (a  2d )  ...  a  (n  1)d
Sn  a  (n  1)d  ...  a
2Sn  2a  (n  1)d  ...  2a  (n  1)d
2Sn  n  2a  (n  1)d 
Sn  12 n  2a  (n  1)d 
And so we have
Sn  12 n  2a  (n 1)d 
A variation on this formula is obtained by substituting the expression for the last term
l  a  (n  1)d . This gives us
Sn  12 n(a  l )
We can make sense of this formula by realising that n is the number of terms, and 12 (a  l ) is the
‘average’ term of the sequence.
An important special case is the formula for the sum of the first n natural numbers (these are the
positive integers). These form an A.P. with n terms, first term 1, and last term n. Substituting into
the formula above,
sum of the first n natural numbers  12 n(n  1)
Example 1 : Find the sum of the first 100 terms of the A.P. with first term 6 and common
difference 0.8.
Sn  12 n  2a  (n  1)d 
S100  12 100  2  6  99  0.8
 50  (12  79.2)
 50  91.2
 4560
Example 2 : Evaluate
14
 7r  5
r 1
This as an A.P. with n = 14, a = 12, d = 7 and l = (7 × 14) + 5 = 103. We can therefore use either
of our two formulas for the sum of an A.P.
Sn  12 n(a  l )
S14  12 14(12  103)
 7 115
 805
Example 3 : The sum of the series 5 + 9 + 13 + ... is 1539. How many terms does the series
contain?
S n  12 n  2a  (n  1)d 
1539  12 n  2  5  (n  1)4
1539  12 n(4n  6)
1539  2n 2  3n
2n 2  3n  1539  0
3  32  4  2  1539
2 2
3  111
n
4
n  27, n  28.5
n
The series has 27 terms.
Example 4 : A pole 9 metres long is sawn into several lengths which form an A.P. The longest
piece is 1·8 metres and the shortest piece is 0·2 metres. How many lengths are
there?
Sn  12 n(a  l )
9  12 n(0.2  1.8)
n9
There are nine lengths.
Example 5 : The fourth term of an A.P. is 10 and the sum of the first nine terms is 72. Find the
first term and the sum of the first twenty terms.
The fourth term…
a  3d  10
The sum of the first nine terms…
Sn  12 n  2a  (n  1)d 
72  12  9(2a  8d )
a  4d  8
Subtracting,
d  2
a  (3  2)  10
a  16
Now
Sn  12 n  2a  (n  1)d 
S20  12  20  2 16  19  2
 10  (32  38)
 60
Example 6 : A company offers employees the choice of two salary schemes for a five-year
contract. In Scheme A, they pay £1000 for the first month with monthly increments
of £15 (i.e. £1015 for the second month, £1030 for the third and so on). In Scheme
B, they pay £14000 for the first year, with annual increments of £800. Which
scheme offers employees the greater total pay over five years?
Scheme A is an A.P. with a = 1000, d = 15, n = 60.
Sn  12 n  2a  (n  1)d 
S20  12  60  2 1000  59 15
 30  (2000  885)
 £86550
Scheme B is an A.P. with a = 14000, d = 800, n = 5.
Sn  12 n  2a  (n  1)d 
S5  12  5  2 14000  4  800
 2.5  (28000  3200)
 £78000
Scheme A offers the greater total pay.
Example 7 : Find the sum of all the integers from 34 to 71 inclusive.
Using the formula for the sum of the first n positive integers,
Sn  12 n(n  1)
S71  12  71 72
 2556
S33  12  33  34
 561
S71  S33  2556  561
 1995
Example 8 : Find the sum of all the multiples of 3 from 1 to 200. Hence find the sum of all the
numbers which are not multiples of 3, from 1 to 200.
200
 66
The multiples of 3 form an A.P. with a = 3, d = 3, n 
3
Sn  12 n  2a  (n  1)d 
S66  12  66  2  3  65  3
 33  (6  195)
 6633
To find the sum of all the non-multiples of 3, we sum all the integers from 1 to 200, and subtract
our previous answer. The sum of the integers from 1 to 200 is given by
Sn  12 n(n  1)
S200  12  200  201
 20100
The non-multiples of 3 add up to 20100 – 6633 = 13467.
Example 9 : The sum of the first n terms of a sequence is given by Sn  2n 2  3n . Prove the
sequence is an A.P., and find its first term and its common difference.
We manipulate the expression for S n until it is in the form 12 n  2a  (n  1)d  .
2n 2  3n  12 n(4n  6)
 12 n  4(n  1)  10
 12 n  2  5  (n  1)4
Comparing with the formula for the sum on an A.P., we see that the sequence must be an A.P.,
and has a = 5 and d = 4.
An alternative method is to realise that the nth term un is given by
un  S n  S n 1
  2n 2  3n    2( n  1) 2  3( n  1) 
  2n 2  3n    2n 2  4n  2  3n  3 
  2n 2  3n    2n 2  n  1
 4n  1
 5  4(n  1)
Comparing with the nth term of an A.P., which is a  (n  1)d , we see our sequence is indeed an
A.P. with first term 5 and common difference 4.
C1 p96 Ex 6F, p98 Ex 6G, p96 Ex 6H
Topic Review : Sequences and Arithmetic Series