Download sequences_and_series_a_and_as_level_mathematics and more Summaries Algebra in PDF only on Docsity!
Sequences
and
series
ASIAN SAVINGS
θ
DOUBLE
your $$ every
10 years
Sequences and series
P
Population, when unchecked, increases in a geometrical
ratio. Subsistence increases only in an arithmetical ratio. A
slight acquaintance with numbers will show the immensity of
the first power in comparison with the second.
Thomas Malthus (1798)
? Each of the following sequences is related to one of the pictures above.
(i) 5000, 10 000, 20 000, 40 000, ….
(ii) 8, 0, 10, 10, 10, 10, 12, 8, 0, ….
(iii) 5, 3.5, 0, –3.5, –5, –3.5, 0, 3.5, 5, 3.5, ….
(iv) 20, 40, 60, 80, 100, ….
(a) Identify which sequence goes with which picture.
(b) Give the next few numbers in each sequence.
(c) Describe the pattern of the numbers in each case.
(d) Decide whether the sequence will go on for ever, or come to a stop.
75
This version has the advantage that the right-hand side begins with the first term of the sequence.
Arithmetic progressions
P
Figure 3.
Any ordered set of numbers, like the scores of this golfer on an 18-hole round
(see figure 3.1) form a sequence. In mathematics, we are particularly interested
in those which have a well-defined pattern, often in the form of an algebraic
formula linking the terms. The sequences you met at the start of this chapter
show various types of pattern.
A sequence in which the terms increase by the addition of a fixed amount
(or decrease by the subtraction of a fixed amount), is described as arithmetic.
The increase from one term to the next is called the common difference.
Thus the sequence 5 8 11 14… is arithmetic with
3 3 3 common
difference 3.
This sequence can be written algebraically as
u
k
= 2 + 3 k for k = 1, 2, 3, …
When k 1, u
1
k 2, u
2
k 3, u
3
and so on.
(An equivalent way of writing this is u
k
5 3( k 1) for k 1, 2, 3, … .)
As successive terms of an arithmetic sequence increase (or decrease) by a fixed
amount called the common difference, d , you can define each term in the
sequence in relation to the previous term:
u
k +
u
k
d.
When the terms of an arithmetic sequence are added together, the sum is called
an arithmetic progression , often abbreviated to A.P. An alternative name is an
arithmetic series.
77
progressions Arithmetic
The 7th term is the 1st term (5) plus six times the common difference (2).
Notation
When describing arithmetic progressions and sequences in this book, the
following conventions will be used:
● first term, u
1
= a
● number of terms = n
● last term, u
n
= l
● common difference = d
● the general term, u
k
, is that in position k (i.e. the k th term).
Thus in the arithmetic sequence 5, 7, 9, 11, 13, 15, 17,
a 5, l 17, d 2 and n 7. The
terms are formed as follows.
u
1
a 5
u
2
a d 5 2 7
u
3
a 2 d 5 2 2 9
u
4
a 3 d 5 3 2 11
u
5
a 4 d 5 4 2 13
u
6
a 5 d 5 5 2 15
u
7
a 6 d 5 6 2 17
You can see that any term is given by the first term plus a number of
differences. The number of differences is, in each case, one less than the number
of the term.
You can express this mathematically as
u
k
a ( k 1) d.
For the last term, this becomes
l a ( n 1) d.
These are both general formulae which apply to any arithmetic sequence.
Find the 17th term in the arithmetic sequence 12, 9, 6, ….
SOLUTION
In this case a 12 and d 3.
Using u
k
a ( k 1) d , you obtain
u
17
78
The 17th term is 36.
EXAMPLE 3.
Sequences
and
series
P
n
4
2
2
Find the value of 8 6 4 … (32).
SOLUTION
This is an arithmetic progression, with common difference 2. The number of
terms, n , may be calculated using
n
l – a
d
n
The sum S of the progression is then found as follows.
S 8 6 … 30 32
S 32 – 30 … 6 8 2 S
Since there are 21 terms, this gives 2 S 24 21, so S 12 21 252.
Generalising this method by writing the series in the conventional notation
gives:
S
n
[ a ] [ a + d ] … [ a ( n 2) d ] [ a ( n 1) d ]
S
n
[ a ( n 1) d ] [ a + ( n 2) d ] … [ a + d ] [ a ]
2 S
n
[2 a ( n 1) d ] [2 a ( n 1) d ] … [2 a ( n 1) d ] [2 a + ( n 1) d ] Since
there are n terms, it follows that
S
n
n
2 a
n
d
This result may also be written as
S
n ( a l ).
Find the sum of the first 100 terms of the progression
1
1
3
4 2 4
SOLUTION
In this arithmetic progression
a 1, d
1
and n 100.
Using S
n
1
n
2 a
( n – 1) d
, you have
S
n
1
1
2 4
1
80
EXAMPLE 3.
EXAMPLE 3.
Sequences
and
series
P
2
2
Jamila starts a part-time job on a salary of $9000 per year, and this increases by
an annual increment of $1000. Assuming that, apart from the increment,
Jamila’s
salary does not increase, find
(i) her salary in the 12th year
(ii) the length of time she has been working when her total earnings are $100 000.
SOLUTION
Jamila’s annual salaries (in dollars) form the arithmetic
sequence 9000, 10 000, 11 000, ….
with first term a 9000, and common difference d 1000.
(i) Her salary in the 12th year is calculated using:
u
k
a ( k 1) d
u
12
(ii) The number of years that have elapsed when her total earnings are $100 000
is given by:
S 1 n
2 a ( n – 1) d
where S 100 000, a 9000 and d 1000.
This gives 100 000
1
n
2 9000 1000( n – 1) .
This simplifies to the quadratic equation:
n
2
17 n 200 0.
Factorising,
( n 8)( n 25) 0
n 8 or n 25.
The root n 25 is irrelevant, so the answer is n 8.
Jamila has earned a total of $100 000 after eight years.
1 Are the following sequences arithmetic?
If so, state the common difference and the seventh term.
(i) 27, 29, 31, 33, … (ii) 1, 2, 3, 5, 8, … (iii) 2, 4, 8, 16, …
(iv) 3, 7, 11, 15, … (v) 8, 6, 4, 2, …
2 The first term of an arithmetic sequence is 8 and the common difference is 3.
(i) Find the seventh term of the sequence.
(ii) The last term of the sequence is 100.
How many terms are there in the sequence?
81
EXERCISE 3A
EXAMPLE 3.
Exercise
3A
P
6300? [MEI]
12 Paul’s starting salary in a company is $14 000 and during the time he
stays with the company it increases by $500 each year.
(i) What is his salary in his sixth year?
(ii) How many years has Paul been working for the company when his total
earnings for all his years there are $126 000?
13 A jogger is training for a 10 km charity run. He starts with a run of 400
m; then he increases the distance he runs by 200 m each day.
(i) How many days does it take the jogger to reach a distance of 10
km in training?
(ii) What total distance will he have run in training by then?
14 A piece of string 10 m long is to be cut into pieces, so that the lengths of
the pieces form an arithmetic sequence.
(i) The lengths of the longest and shortest pieces are 1 m and 25 cm
respectively; how many pieces are there?
(ii) If the same string had been cut into 20 pieces with lengths that
formed an arithmetic sequence, and if the length of the second longest
had been
92.5 cm, how long would the shortest piece have been?
15 The 11th term of an arithmetic progression is 25 and the sum of the first
4 terms is 49.
(i) Find the first term of the progression and the common
difference. The n th term of the progression is 49.
(ii) Find the value of n.
16 The first term of an arithmetic progression is 6 and the fifth term is 12. The
progression has n terms and the sum of all the terms is 90. Find the value of
n.
[Cambridge AS & A Level Mathematics 9709, Paper 1 Q3 November 2008]
17 The training programme of a pilot requires him to fly ‘circuits’ of an airfield.
Each day he flies 3 more circuits than the day before. On the fifth day he
flew 14 circuits.
Calculate how many circuits he flew:
(i) on the first day
(ii) in total by the end of the fifth day
(iii) in total by the end of the n th day
(iv) in total from the end of the n th day to the end of the 2 n th day.
Simplify your answer.
[MEI]
Exercise
3A
P
metric
progressions
3
P
18 As part of a fund-raising campaign, I have been given some books of raffle tickets to sell. Each book has the same number of tickets and all the tickets I have been given are numbered in sequence. The number of the ticket on the front of the 5th book is 205 and that on the front of the 19th book is 373. (i) By writing the number of the ticket on the front of the first book as a and the number of tickets in each book as d , write down two equations involving a and d. (ii) From these two equations find how many tickets are in each book and the number on the front of the first book I have been given. (iii) The last ticket I have been given is numbered
- How many books have I been given?
[MEI]
Geo
Figure 3. A human being begins life as one cell, which divides into two, then four…. The terms of a geometric sequence are formed by multiplying one term by a fixed number, the common ratio, to obtain the next. This can be written inductively as: u k + ru k with first term u 1
The sum of the terms of a geometric sequence is called a geometric progression , shortened to G.P. An alternative name is a geometric series. Notation When describing geometric sequences in this book, the following conventions are used: ● first term u 1 = a 84 ● common ratio = r Sequences and series
5
1 – 2 – = 3
4
● number of terms = n
● the general term u
k
is that in position k (i.e. the k th term).
Thus in the geometric sequence 3, 6, 12, 24, 48,
a 3, r 2 and n 5.
The terms of this sequence are formed as follows.
u
1
a 3
u
2
a r 3 2 6
u
3
a r
2
2
u
4
a r
3
3
u
5
a r
4
4
You will see that in each case the power of r is one less than the number of the
term: u
5
ar
4
and 4 is one less than 5. This can be written deductively as
u
k
ar
k –
, and
the last term is
u
n
ar
n –
These are both general formulae which apply to any geometric sequence.
Given two consecutive terms of a geometric sequence, you can always find
the common ratio by dividing the later term by the earlier. For example,
the geometric sequence … 5, 8, … has common ratio r
8
Find the seventh term in the geometric sequence 8, 24, 72, 216, ….
SOLUTION
In the sequence, the first term a 8 and the common ratio r 3.
The k th term of a geometric sequence is given by u
k
ar
k –
and so u
7
6
How many terms are there in the geometric sequence 4, 12, 36, … , 708 588?
SOLUTION
Since it is a geometric sequence and the first two terms are 4 and 12, you can
immediately write down
First term: a = 4
Common ratio: r = 3
85
EXAMPLE 3.
EXAMPLE 3.
progressions Geometric
P
Now multiply it by the common ratio, 2:
2 S 2 4 8 16 … 2
64
2
Then subtract
1 from
2
2 2 S 2 4 8 16 … 2
63
64
1 S 1 2 4 8 … 2
63
subtracting: S –1 0 0 0 … 2
64
The total number of rice grains requested was therefore 2
64
1 (which is about
19
? How many tonnes of rice is this, and how many tonnes would you expect there
to be in China at any time?
(One hundred grains of rice weigh about 2 grammes. The world
annual production of all cereals is about 1.8 10
9
tonnes.)
Note
The method shown above can be used to sum any geometric progression.
Find the value of 0.2 1 5 … 390 625.
SOLUTION
This is a geometric progression with common ratio 5.
Let S 0.2 1 5 … 390 625.
1
Multiplying by the common ratio, 5, gives:
5 S 1 5 25 … 390 625 1 953 125.
2
Subtracting
1
from
2
:
5 S 1 5 25 … 390 625 1 953 125
S 0.2 1 5 25 … 390 625
4 S 0.2 0 … 0 1 953 125
This gives 4 S 1 953 124.
S 488 281.2.
87
EXAMPLE 3.
progressions Geometric
P
2 2 2 2
3
P
The same method can be applied to the general geometric progression to give a formula for its value:
S
n a ar ar 2 … ar n –
1 Multiplying by the common ratio, r , gives: rS n ar ar 2 ar 3 … ar n
2 Subtracting
1 from
2 , as before, gives: ( r 1) S n – a ar n a ( r n
so S n
a ( r n
( r 1) This can also be written as:
S
a (1 – r n
n (1 – r ) Infinite geometric progressions The progression
1
1
1
1
is geometric, with common ratio 1
2 4 8 16 2 Summing the terms one by one gives 1, 1 1
3
7
15
2 4 8 16 Clearly the more terms you take, the nearer the sum gets to 2. In the limit, as the number of terms tends to infinity, the sum tends to 2. As n ∞, S n
This is an example of a convergent series. The sum to infinity is a finite number. You can see this by substituting a 1 and r 1 in the formula for the sum of the series:
S
n
a 1 – r
n
1 – r
1 n giving
S
n
1
1 n
The larger the number of terms, n , the smaller 1 n becomes and so the nearer S is to the limiting value of 2 (see figure 3.3). Notice that 2 can never be negative, 2 n n 1 however large n becomes; so S n can never exceed 2. 88 Sequences and series
89
9 4 4 ∞
3 3 3
3
P
Note ˙ You may have noticed that the sum of the series 0.2 + 0.02 + 0.002 + … is 0.2, and that this recurring decimal is indeed the same as 2 . The first three terms of an infinite geometric progression are 16, 12 and 9. (i) Write down the common ratio. (ii) Find the sum of the terms of the progression. SOLUTION (i) The common ratio is 3
(ii) The sum of the terms of an infinite geometric progression is given by:
S
a 1 – r In this case a = 16 and r = 3 , so:
S
∞
4
? A paradox Consider the following arguments. (i) S 1 2 4 8 16 32 64 …
S 1 2(1 2 4 8 16 32 …)
1 2 S
3 S 1
S
1
(ii) S 1 ( 2 4) ( 8 16) ( 32 64) …
S 1 2 8 32 … So S
diverges towards ∞. (iii) S (1 2) (4 8) (16 32) …
S –1 4 8 16 …
So S diverges towards ∞. What is the sum of the series: 1 , ∞, ∞, or something else? 90 EXAMPLE 3. Sequences and series
