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© Boardworks Ltd 2005 of 68 A7 Sequences KS4 Mathematics

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KS4 Mathematics. A7 Sequences. A7 Sequences. Contents. A. A7.2 Linear sequences. A. A7.1 Generating sequences from rules. A7.3 Quadratic sequences. A. A7.4 Geometric sequences. A. A7.5 Other types of sequence. A. Sequences. 1 st term. 6 th term. - PowerPoint PPT Presentation

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A7 SequencesA
A
A
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© Boardworks Ltd 2005
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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.
Sequences
1st term
6th term
Explain that it is important to know the position of terms in a sequence because they are often used to define the sequence.
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Predicting terms in a sequence
Sometimes, we can predict how a sequence will continue by looking for patterns.
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.
67
60
What are the next two terms in the following sequence,
102, 95, 88, 81, 74 . . . ?
–7
–7
–7
–7
102
95
88
81
74
–7
–7
The first step in determining the nature of a sequence is to find the difference between each term. This will determine whether or not the sequence is linear.
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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?
14
22
32
44
16
32
64
128
+2
+4
2
4
8
+6
+8
+10
+12
×2
×2
2
4
8
×2
×2
×2
×2
Ask pupils if they can think of any other ways that this sequence could continue. For example, we could add 2, add 4, add 8, add 16, etc.
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Defining sequences
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.
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.
The first is to use a term-to-term rule.
The second is to use a position-to-term rule.
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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
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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.
1
4
9
16
25
Position
Term
1st
2nd
3rd
4th
5th
nth
n2
Once we know a position-to-term rule we can find any term in the sequence given its position in the sequence.
Ask pupils to give other terms in the sequence with the position-to-term rule n2.
For example, what is the 12th term in the sequence?
You could also ask pupils to give you the position of a given term in the sequence using inverse operations. For example,
81 is a term is this sequence. What position is it in?
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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
u1 =
15
The first five terms in the sequence are: –1, 3, 7, 11 and 15.
Writing sequences from position-to-term rules
Ask pupils what they notice about this sequence. (It goes up 4 each time. An even better answer is: it is the numbers from the 4 times table with 5 subtracted each time.)
Ask pupils to use the rule to work out the value of other terms in the sequence.
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We can write this rule as:
un = 2n2 + 3
u1 =
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.
In a linear sequence the n can only be raised to the power of 1 (though this is not usually written). In a quadratic sequence, the rule for the nth term involves n raised to the power of 2 and possibly n raised to the power of 1 (but no higher) as well.
This is discussed in detail in A7.3 Quadratic sequences.
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Writing sequences from position-to-term rules
Ask pupils to calculate the value of each term using substitution. Calculators may be used for more difficult examples.
If the example shown is of a form you don’t want, press the reset button for another example.
1199.unknown
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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, . . .
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.
+6
+6
+6
+6
+6
+6
+6
Remind pupils that the sequence of multiples of 6 increases by adding 6 each time.
Ask pupils how this sequence is related to multiples 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, . . .
When the difference between each term in a sequence is always the same we have a linear or arithmetic sequence.
–3
–3
–3
–3
–3
–3
–3
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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
× 3
× 3
× 3
× 3
× 3
× 3
+ 1
+ 1
+ 1
+ 1
+ 1
+ 1
Explain that this sequence is related to multiples of 3 because, like the sequence of numbers in the 3 times table, we find each term by adding 3 to the previous term.
Encourage pupils to verify the formula for the nth term by substitution. For example, using the formula, the 4th term should be 3 × 4 + 1 = 12 + 1 = 13 and the 5th term should be 3 × 5 + 1 = 15 + 1 = 16. These are true and so we can assume that our formula is correct.
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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
× –2
× –2
× –2
× –2
× –2
× –2
+ 7
+ 7
+ 7
+ 7
+ 7
+ 7
Explain that it is preferable not to have a negative sign at the beginning of an expression. We therefore write –2n + 7 as 7 – 2n.
Again, encourage pupils to verify the formula for the nth term by substitution. For example, using the formula, the 4th term should be 7 – 2 × 4 = 7 – 8 = –1 and the 5th term should be 7 – 2 × 5 = 7 – 10 = –3. These are true and so we can assume that our formula is correct.
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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,
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.
Use this slide to explain that the nth term of any linear sequence can be found by multiplying n by the common difference d and adding the difference between the first term a and the common difference d.
Explain that we can remember this general form by thinking of (a – d) as the 0th term. This term does not really exist but can easily be found by using the common difference to find the term that would come before the first term. This value can be compared to the y-intercept on a graph.
The general form is always ‘the common difference × n minus the 0th term’.
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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
Position number (n)
Value of term (un)
Stress that unlike the graph of a function, the intermediate values on the line have no meaning because the position number n can only be a whole number. There’s no such thing as the 4.5th point, for example.
We can use the gradient of the graph to tell us the difference between the terms and we can use the y-intercept to tell us the value of the 0th term.
Establish that if the gradient of the graph is 2 and the y-intercept is 1, the nth term of the sequence is un = 2n + 1.
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Finding the nth term of a linear sequence
Use this activity to review finding the nth term of a linear sequence.
Reveal any three consecutive terms and ask pupils to find the remaining terms. By finding the common difference, or otherwise, ask pupils to find the rule for the nth term.
Make the activity more difficult by revealing three non-consecutive terms.
The value of the 0th term has been included so that the link between its value and the rule for the nth term can be established.
Once the rule has been revealed check by substitution that all of the terms fit the rule.
1201.unknown
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,
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.
+4
+4
+4
+4
Point out that analyzing the sequence alone only allows us to predict how many tiles there will be in the next pattern. We can only prove that this is true by drawing the next pattern and counting the tiles or by justifying the sequence in the context of the 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,
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.
+4
+4
+4
+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
un = 4n – 3.
We can use this formula to predict the number of tiles in the 20th pattern:
u20 = 4 × 20 – 3
– 3
– 3
– 3
– 3
– 3
– 3
We can also find the value of c by working out the value of the first term minus the common difference (the 0th term).
Explain that we can only know for sure that the 20th pattern will contain 77 tiles by drawing it or by justifying that the formula is true using the original context. When we are considering problems arising from practical contexts, using the sequence alone is not enough.
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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, . . .
This sequence starts with 4 and increases by adding consecutive whole numbers to each term.
The differences between the terms form a linear sequence.
+1
+2
+3
+4
+5
+6
+7
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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, . . .
With sequences of this type it is often helpful to find a second row of differences.
–0.1
–0.2
–0.3
–0.4
–0.5
–0.6
–0.7
Explain that a second row of differences shows the differences between the differences. In this example, the difference between each difference is 0.1.
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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
A sequence is formed by the first row of differences so we look at the second row of differences.
This shows that the differences increase by 3 each time.
First row of differences
Second row of differences
+2
+5
+8
+11
+14
+17
+20
+3
+3
+3
+3
+3
+3
Point out to pupils that the first row of differences forms a linear sequence with a common difference of 3.
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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
Investigating quadratic sequences
Use this activity to practice substituting position numbers into the general form of a quadratic sequence.
Investigate the connection between the constant term in the second row of differences and the coefficient of n2 in the general term. Establish that the second difference will always be equal to 2a, where a is the coefficient of n2.
Investigate the connection between the value of the 0th term and the value of c in the general form an2 + bn + c.
1202.unknown
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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
Explain that to find the nth term of any given quadratic sequence we have to find the values of a, b and c for in general form an2 + bn + c.
Talk through how we can find the first 5 terms of a general quadratic sequence by substituting the position number for n in an2 + bn + 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
This shows that the second difference is always 2a when
un = an2 + bn + c.
Stress that all quadratic sequences will be of this form for any given values of a, b and c where a ≠ 0.
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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
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.
This rule is not as easy to remember as the previous two rules. It is usually best to find b within the context of the problem.
You can tell pupils that they are unlikely to be asked to find the nth term of a quadratic sequence in an exam. However, it is helpful to know these techniques because they often come up when working on coursework investigations.
© Boardworks Ltd 2005
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
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
+5
+9
+13
+17
+4
+4
+4
We start by finding the value of a in un = an2 + bn + c by considering the value of the second differences.
© Boardworks Ltd 2005
The nth term of a quadratic sequence
Let’s start by looking at the first and second differences.
4, 9, 18, 31, 48
The value of c is the same as the value for the 0th term.
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, …
+5
+9
+13
+17
+4
+4
+4
+4
+1
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
4 = b + 5
un = 2n2 + bn + 3
b = –1
Find the nth term of the sequence, 4, 9, 18, 31, 48, …
More generally u1 = a + b + c so b = u1 – (a + c).
In this example u1 is 4, a is 2 and c is 3 so, b = 4 – 5 = –1.
© Boardworks Ltd 2005
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
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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
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
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
The value of c is the same as the value for the 0th term.
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
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
1 = ½ + b
un = ½n2 + bn
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)
n2
2
n
2
un =
Check the rule for other terms in the sequence.
Explain that there are many ways to express the general term for a triangular number. For example, factorizing would give un = ½n(n + 1). The way it is expressed generally depends on the context from which it is generated.
An alternative derivation of this formula is given in N1.1 Classifying numbers.
Encourage pupils to remember this rule because triangular numbers appear often in mathematical investigations.
© Boardworks Ltd 2005
Finding the nth term of a quadratic sequence
Use this activity to randomly generate quadratic sequences by clicking on the blue arrows.
Reveal the first five terms in the sequence and ask pupils how we can proceed using differences and the value of the 0th term to find the formula for the nth term.
1203.unknown
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
13 tiles
Pattern 4
25 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, 13, 25,
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.
+4
+8
+12
+16
+4
+4
+4
Point out that analyzing the sequence alone only allows us to predict how many tiles there will be in the next pattern. We can only prove that this is true by drawing the next pattern and counting the tiles or by justifying the sequence in the context of the pattern.
© Boardworks Ltd 2005
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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:
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,
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
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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,
41
If a = 2 and c = 1 then un = 2n2 + bn + 1,
Look at the pattern made by the differences,
u1 = 2 + b + 1
+4
+8
+12
+16
+4
+4
+4
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
= 800 – 40 + 1
= 761
We can predict that there will be 761 tiles in the 20th pattern.
Again, unless we can prove that this formula is true from the context of the original pattern, we can only predict the number of tiles in the 20th pattern.
As an extension activity pupils could be asked to justify the formula for the nth term by analysing how the original pattern is built up. For example by using alternate colours pupils should be able to see that each pattern can be found by adding n2 and (n – 1)2. This is equivalent to 2n2 – 2n + 1.
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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, . . .
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.
×2
×2
×2
×2
×2
×2
×2
Point out that this sequence increases in unequal steps. It is not linear.
If we looked at the differences between the consecutive terms they would be 2, 4, 8, 16, 32 … in other words, they would form the same sequence.
This is a sequence of powers of 2. We could write it as 21, 22, 23, 24, 25, 26, 27 …
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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
1
4
We could also continue this sequence by multiplying each term by to give the next term.
1
16
,
Each term in this sequence is one quarter of the term before. Explain that multiplying by ¼ is equivalent to dividing by 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, …
The sequence continues by multiplying the previous term by 1.5.
r = 12 ÷ 8 =
1.5
We can check that a sequence is geometric by dividing each term by the previous term. If the answer is always the same then it is geometric.
We can think of the terms is the example as increasing by 50% each time.
© Boardworks Ltd 2005
Continuing geometric sequences
Start by finding the common ratio by dividing any term by the one before it. Use this to find the next two terms.
None of the common ratios in these examples are negative. Pupils could investigate the effect of having a negative common ratio as an extension activity.
1204.unknown
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
u4 = 24 = 3 × 2 × 2 × 2 = 3 × 23
u5 = 24 = 3 × 2 × 2 × 2 × 2 = 3 × 24
un = 3 × 2(n – 1)
Explain how this sequence can be written by multiplying the first term by 2 raised to successive powers.
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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,
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,
£840,
£882,
£926.10, …
×1.05
×1.05
×1.05
See N5.6 Compound percentages.
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
un = £800 × 1.05n – 1
Point out that if the formula was showing the value of the investment at the end of the first year rather than the beginning, the formula for un would be £800 × 1.05n.
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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
6
9
12
15
18
21
2
3
Find the formula for the nth term of the sequence.
The sequence formed by the numerators is:
u1 = 2
u2 = 3
u3 = 4
u1 = 3
u2 = 6
u3 = 9
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:
2
3
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,
This sequence is called the Fibonacci Sequence after the Italian mathematician who first wrote about it.
21,
34,
55, . . .
Sequences of this type must be generated by two numbers.
Ask pupils if this type of sequence can be descending. (If the sequence is generated by two negative numbers then it will be a descending sequence.)
The Fibonacci Sequence appears in many situations in nature. Ask pupils to research some examples on the Internet.
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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:
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!
Explain that iterative formulae express terms in a sequence using previous terms. They are a good way of expressing term-to-term rules algebraically where a position-to-term rule is difficult or impossible to find.
The disadvantage of iterative formulae is that to find any given term in the sequence you have to work out every term from the beginning.
Pupils could investigate Fibonacci-type sequences with different generating numbers.
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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, …
This is a case where we can generate a position-to-term rule, knowing the iterative rule. Talk through how the sequence is generated.
The iterative formula is basically a position-to-term rule. In this example, each term is double the one before it plus 1.