© boardworks ltd 2005 1 of 68 a7 sequences ks4 mathematics

© Boardworks Ltd 2005 1 of 68

A7 Sequences

KS4 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


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,


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


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


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.


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


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

n21 4 9 16 25


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.


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.



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.


This sequence is a quadratic sequence.


Matching sequences to rules


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A7 Sequences



A7.1 Generating sequences from rules



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.


Can you work out the next three terms in this sequence?

How did you work these out?

This sequence starts with 11 and decreases by subtracting 3 from each term to give the next term.

Sequences that decrease in equal steps

11, 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.

When the difference between each term in a sequence is always the same we have a linear or arithmetic sequence.


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.


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.un = 3n + 1.


Find the nth term of the sequence 5, 3, 1, –1, –3, …


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.un = 7 – 2n.


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:


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)un = dn + (a – d)(a – d) is the value

of the 0th term.


Position number (n)

Val

ue o

f te

rm (

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 + 1un = 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


Finding the nth term of a linear sequence


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 39 tiles Pattern 4

13 tiles


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.


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.


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.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 Sequences






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.




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.



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


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


Investigating quadratic sequences


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.

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



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.


When un = an2 + bn + c the value of c can be found by calculating the value of the 0th term, u0.

When un = an2 + bn + c the value of c can be found by calculating the value of the 0th term, u0.



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 u0 = c


When un = an2 + bn + c the value of b can be found by subtracting (a + c) from the value of the first term.

When un = an2 + bn + c the value of b can be found by subtracting (a + c) from the value of the first term.



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)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.



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 = 2a = 2



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 = 3c = 3




Putting a = 2 and c = 3 into un =an2 + bn + c gives usun = 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




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 + 3un = 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

= 48Check the rule for other terms in the sequence.




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 = ½a = ½




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 = 0c = 0




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 = ½b = ½



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+

n2

or

un =n2 + n

2


Finding the nth term of a quadratic sequence


Tiling patterns

The following patterns are made from tiles,

Pattern 1

1 tile Pattern 2

5 tiles Pattern 313 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?


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.


Tiling patterns


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


Tiling patterns


1, 5, 13, 25,

+4 +8 +12 +16

+4 +4 +4

41

The value of c is equal to the 0th term.


The 0th term is equal to 1, so:

c = 1

+4

+0

1,


Tiling patterns


1, 5, 13, 25,

+4 +8 +12 +16

+4 +4 +4

41

If a = 2 and c = 1 then un = 2n2 + bn + 1,


u1 = 2 + b + 1

u1 = b + 3

1 = b + 3

b = –2

1 is the first term.


Tiling patterns

We can now substitute a = 2b = –2c = 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 + 1un = 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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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.




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

, . . .116

14 ,


Geometric sequences

Sequences that go from one term to the next by multiplying by a constant amount are called 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 = 1.5The sequence continues by multiplying the previous term by 1.5.


Continuing geometric sequences


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:


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)un = 3 × 2(n – 1)


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 – 1un = arn – 1

u5 = a × r4,


Geometric sequences in real-life

Geometric sequence 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, …


Geometric sequences in real-life

Geometric sequence 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 = £800u2 = £800 × 1.05u3 = £800 × 1.052

u4 = £800 × 1.053

un = £800 × 1.05n – 1


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A7 Sequences






Fraction sequences

The terms in the following sequence are all fractions:

23

, 36

, 49

, 512

, 615

, …

Find the next two terms in the sequence.

Look at the sequence formed by the numerators and the sequence formed by the denominators separately.

32 ,

63 ,

94 ,

125 ,

156 ,

+3 +3 +3 +3 +3 +3

+1 +1 +1 +1 +1 +1

187 ,

218 , …


Fraction sequences

The terms in the following sequence are all fractions,

23

, 36

, 49

, 512

, 615

, …

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


Fraction sequences

The terms in the following sequence are all fractions:

23

, 36

, 49

, 512

, 615

, …

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


The Fibonacci Sequence



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


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 un = un – 1 + un – 2


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 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!


Using iterative formulae to generate sequences

Write down the first five terms of the sequence generated by the iterative formula

un = 2un – 1 + 1when 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, …

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