Recurrence Relations, Arithmetic and Geometric Sequence

Year 12 Mathematics Applications

Recurrence Relations & Arithmetic and Geometric Sequence

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Contents

Signposts… ...... 2

Overview …...... 3

Lesson 1 – Patterns and relationships ...... 5

Lesson 2 – Patterns and relationships ...... 12

Lesson 3 – Rules for sequences ...... 18

Lesson 4 – Difference equations ...... 25

Lesson 5 – Difference equations ...... 31

Lesson 6 – Arithmetic sequences ...... 36

Lesson 7 – Arithmetic sequences ...... 41

Lesson 8 – Geometric sequences ...... 48

Lesson 9 – Geometric sequences ...... 53

Lesson 10 – Application of sequences ...... 58

Lesson 11 – Exam Practice – Calculator Free ...... 65

Lesson 12 – Exam Practice – Calculator allowed ...... 68

Further practice ...... 72

Glossary ...... 72 Summary ...... 74 Solutions ...... 76

Signposts

Each symbol is a sign to help you. Here is what each one means.

Important Information

Mark and Correct your work

You write an answer or response

Use your CAS calculator

A point of emphasis

Refer to a text book

Contact your school teacher (if you can)

Check your school about Assessment submission

Overview

This Booklet contains approximately 4 weeks of work. You may find it necessary to do ‘homework’ in order to finish it.

To guide the pace at which you work through the booklet, refer to the contents page.

Space is provided for you to write your solutions in this PDF booklet. If you need more space, then attach a page to the page you are working on.

Answers are given to all questions. It is assumed you will use them responsibly, to maximise your learning. You should check your day to day lesson work.

Assessments

All of your assessments are provided for you separately by your school.

Assessments will be either response or investigative. Weightings for assessments are provided by your school.

Calculator

This course assumes the use of a CAS calculator. Screen displays will appear throughout the booklets to help you with your understanding of the lessons. Further support documents are available.

Textbook

You are encouraged to use a text for this course. A text will further explain some topics and can provide you with extra practice questions.

Online Support

Search for a range of online support.

Content covered in this booklet

The syllabus content focused on in this booklet includes:

The arithmetic sequence 3.2.1 use recursion to generate an arithmetic sequence 3.2.2 display the terms of an arithmetic sequence in both tabular and graphical form and demonstrate that arithmetic sequences can be used to model linear growth and decay in discrete situations 3.2.3 deduce a rule for the term of a particular arithmetic sequence from the pattern of the terms in 𝑡𝑡anℎ arithmetic sequence, and use this rule to make 𝑛𝑛 predictions 3.2.4 use arithmetic sequences to model and analyse practical situations involving linear growth or decay The geometric sequence 3.2.5 use recursion to generate a geometric sequence 3.2.6 display the terms of a geometric sequence in both tabular and graphical form and demonstrate that geometric sequences can be used to model exponential growth and decay in discrete situations

3.2.7 deduce a rule for the term of a particular geometric sequence from the pattern of the terms in 𝑡𝑡theℎ sequence, and use this rule to make predictions 𝑛𝑛 3.2.8 use geometric sequences to model and analyse (numerically, or graphically only) practical problems involving geometric growth

Sequences generated by first-order linear recurrence relations 3.2.9 use a general first-order linear recurrence relation to generate the terms of a sequence and to display it in both tabular and graphical form 3.2.10 generate a sequence defined by a first-order linear recurrence relation that gives long term increasing, decreasing or steady-state solutions 3.2.11 use first-order linear recurrence relations to model and analyse (numerically or graphically only) practical problems

Lesson 1 Patterns and relationships

By the end of this lesson you should be able to: • consider situations in which number patterns are generated • make predictions about number sequences based on observed patterns • devise ways of describing patterns of a sequence by a recurrence relation.

Patterns and Relationships

Patterns and relationships are really what mathematics is all about. Many great discoveries and predictions have been made from patterns being identified and the reasons for them being explored.

Fibonacci is one pattern we will explore. The diagram on the left shows how each Fibonacci number is determined by adding the two previous terms. Fibonacci patterns are found extensively in nature, like the Spiral Aloe plant on the right!

Image: Call Tree for Fibonacci Number, Wikimedia. Image: J Brew, Aloe polyphilla. Wikimedia ,

Sequences

A number sequence is a set of numbers arranged in a definite order.

For example

3, 8, 13, 18, 23,… is a sequence formed by a starting number, 3, and then by adding 5 to generate the next number.

As this sequence continues indefinitely, it is called an infinite sequence.

The sequence 5, 10, 20, 50, 100, 200 is the sequence of values (in cents) of Australian coins.

This sequence is a finite sequence because it has a definite number of terms.

Look at the sequence 3, 8, 13, 18, 23, …

Each number of the sequence is called a term.

3 is the first term i.e. t1

8 is the second term i.e. t2

13 is the third term i.e. t3 etc

Note: The general term for a sequence is usually given the symbol tn , but this

can be either lower case or upper case (capital letters), so Tn is equally acceptable. The subscript, n, tells us the term number.

Skills Development 1.1

1. For the sequence 4, 7, 10, 13, 16, 19, … determine

− (a) t4 (b) tt32

+ − (c) 3tt34 (d) 2 (tt54)

2. For the sequence 1, 3, 6, 10, 15, 21, 28, … determine

− (a) T3 (b) 3TT12

− − (c) TT56 (d) TT14(2 T 3)

Determining particular terms of a sequence

If tn is any particular term

then tn +1 is the next term in the sequence (the term one on from tn )

and tn +2 is the term following tn +1 or the term two on from tn

Similarly

If tn is any particular term

then tn −1 is the term before tn or the previous term of the sequence

and tn −2 is the term before tn −1 or two terms before tn

Using symbols, the sequence looks like this:

… , tn −2 , tn −1 , tn , tn +1 , tn +2 , …

Example

Take the sequence 3, 8, 13, 18, 23, …

= If we choose tn 13 then find:

(a) tn +1 (b) tn +2 (c) tn −1 (d) tn −2

Solution

(a) tn +1 (b) tn +2 (c) tn −1 (d) tn −2

= 18 = 23 = 8 = 3

Example

Take the sequence 3, 8, 13, 18, 23, …

= If we choose tn 18 then find:

(a) tn +1 (b) tn +2 (c) tn −1 (d) tn −2

Solution

(a) tn +1 (b) tn +2 (c) tn −1 (d) tn −2

= 23 = 28 = 13 = 8

Skills Development 1.2

1. For each of the sequences below:

(a) Look at the pattern and find the next two terms. (b) Describe the pattern in words. (c) Complete the statement about the terms.

(i) 5, 9, 13, 17, 21, ___ , ___ (ii) 200, 100, 50, 25, 12.5, ___ , ___

If tn = 17, then If tn −1 = 50, then

tn −1 = ______tn = ______

tn +1 = ______tn +1 = ______(iii) 10, 7, 4, 1, -2, ___ , ___ (iv) -1, 0, 3, 8, 15, 24, ___ , ___

If tn = 4, then If tn +1 = 24, then

tn −1 = ______tn = ______

tn +1 = ______tn −1 = ______(v) 2, 4, 8, 16, 32, ___ , ___ (vi) 2, 3, 5, 9, 17, ___ , ___

If tn = 64, then If tn +2 = 33, then

tn −1 = ______tn = ______

tn +1 = ______tn −1 = ______

Recurrence relations

The rule which tells you how to get from one term to the next is called a recurrence relation.

A sequence is defined recursively (i.e. by a recurrence relation) if

(a) the first, or first few terms are specified, and

(b) the remaining terms are defined in terms of previous ones.

Example

Find the second and third terms for the following recurrence relation:

= + = ≥ ttnn+1 4 , t1 2 for n 1

Solution

= + ttnn+1 4

= If n = 1 (and remember that t1 2 )

then:

= + tt214 =24 + = 6

If n = 2 then:

= + tt324 =64 + = 10

The sequence is 2, 6, 10, …

Example

Determine the first three terms for the sequence defined by:

= − = ≥ ttnn+1 31, t1 2 for n 1

Solution

= − ttnn+1 31

This says that “the next term is obtained by multiplying the previous term by 3 and then subtracting 1”.

If n = 1 then:

= − tt2132 =32( ) − 1 = 5

If n = 2 then:

= − tt3232 =35( ) − 1 = 14

The sequence is 2, 5, 14, …

Remember that

tn is the term We could also have a recursive definition containing tn and tn −1 . following tn −1

Example

Determine the first three terms of the sequence defined by:

= = ≥ ttnn3 −1 , t1 4 for n 2

Solution

= ttnn3 −1

If n = 2 then:

= tt213 = 34( ) = 12

If n = 3 then:

= tt323 = 3( 12) = 36

The sequence is 4, 12, 36, …

Lesson 2 Patterns and relationships

By the end of this lesson you should be able to: • devise ways of describing patterns of a sequence by a recurrence relation.

Writing recursive definitions

It is important to realise that a recurrence relation (recursive definition) has two

parts (the part relating tn and tn +1 and the value of t1) and that neither alone is sufficient.

Image: G Jacquenot, Pentagram recursive, Wikimedia

Note: The recurrence relation could relate any two consecutive terms, such as tn

and tn −1 .

Example

Write a recursive definition involving tn and tn +1 for the sequence:

5, 11, 17, 23, 29, …

Solution

Pattern: The next term is equal to the term before it, plus 6.

= + = So the recursive definition is ttnn+1 6 , where t1 5 .

Example

Write a recursive definition involving tn and tn −1 for the sequence:

3, 10, 31, 94, …

Solution

What is the pattern? Some guess and check is in order.

Pattern: The next term is equal to three times the term before it, plus 1.

= + = So the recursive definition is ttnn31−1 , where t1 3 .

Skills Development 2.1

1. For each of the sequences below,

(i) write a recursive definition to connect tn and tn −1

(ii) write a recursive definition to connect tn and tn +1

(a) 2, 8, 32, 128, …

(b) 100, 50, 25, 12.5, …

(d) 4, 5.5, 7, 8.5, 10, …

2. Find the first 5 terms of the following recursively defined sequences.

= = ≥ (a) ttnn+1 3 t1 3 for n 1

1 = + = ≥ (b) gn +1 1 g1 1 for n 1 gn

2 = = ≥ (c) hhnn+1 ( ) h1 2 for n 1

64 = = ≥ (d) tn +1 t1 4 for n 1 tn

= + = ≥ (e) ttnn−1 3 n t1 4 for n 2

3. A car sales company currently has a 200 cars on its lot. On average, they sell 5% of the cars they have every month. On average they add 12 cars to their stock every month.

Let be the number of cars on the lot after months.

𝑛𝑛 𝐶𝐶 (a) Write down a recurrence relation𝑛𝑛 that can be used to model the number of cars on the lot at the end of each month.

(b) Use the recurrence relation to determine the number of cars in the yard after 4 months, rounded to the nearest whole number.

(c) The yard is at full capacity when there are 220 cars in the yard. At the end of which month is the yard considered full?

4. Stephen breeds and sells crickets to pet stores, who then sells them as food for lizards.

When he starts the business, Stephen has 1500 crickets.

(a) The population of crickets grows by 4% per month. Write a recurrence relation to show the population of crickets at the end of each month.

(b) During which month would the population of crickets first become greater than 2000?

(c) Stephen starts selling crickets at the start of the 7th month. He initially limits sales to a maximum of 400 crickets per month.

Write a new recurrence relation to represent this situation.

(d) Explain, using reasoning, why this is not a good business model.

(e) What is the maximum number of crickets that Stephen should sell each month so as to keep an increasing population?

5. A communications company, VodaTelOpt, currently supplies NBN to 430 homes in a particular suburb. Each month, an average of 18% of their customers cancel their service and an average of 8 new customers sign up to an NBN plan.

Let be the number of customers of VodaTelOpt after months.

𝑉𝑉𝑛𝑛 (a) Write down a recurrence relation that can𝑛𝑛 be used to model the number of customers at the end of each month.

(b) Use the recurrence relation to determine the number of customers of the internet service provider after 4 months, rounded to the nearest whole number.

Lesson 3 Rules for sequences

By the end of this lesson you should be able to: • find the general rule for a sequence • generate a sequence from a general rule.

Consider the sequence 6, 8, 10, 12, 14, …

= = = This sequence has t1 6 , t2 8 , t3 10 , and so on. Let’s put this information into a table.

Note: The term number is in the top row of the table and the value of the term is in the second row of the table.

The first difference pattern has a constant value of 2, so it is a linear equation. This common difference is the multiply factor in the general equation.

= + tn (multiply factor).nc

∴ = + tn 2 nc

The value of c is the value of tn when n = 0. Following the pattern backwards we get:

∴ = + tnn 24

This general rule allows us to generate any term of the sequence.

To find the value of t10 substitute n = 10 into the general equation.

∴ = += t10 2( 10) 4 24

Image: Maths Enhancement Project, Book 9.

This is a linear sequence shown visually.

Example

Determine the general rule for the sequence 4, 7, 10, 13, … and then use

the rule to determine t15 of the sequence.

Solution

The sequence can be continued by adding 3 each time so the constant first = + difference is 3. i.e. tn 3 nc.

= = Take any known point of the sequence. We will use t2 7 , so tn 7 when n = 2.

= + tn 3 nc

You can use your calculator to ∴ 7= 32( ) + c determine rules; however, it is important that you can also determine rules algebraically ∴ c = 1 like shown here.

∴ = + tnn 31

This rule will now allow us to determine the 15th term of the sequence.

∴ = += t15 3( 15) 1 46

An alternative algebraic approach

Sequence is 4, 7, 10, 13, …

As the first difference pattern has a constant value of 3, then this common difference is the multiply factor.

∴ = + tn 3 nc

The value of c is the value of tn when n = 0. Following this pattern back we get:

∴ = + tnn 31

Example

Determine the general rule for the sequence 2, 7, 14, 23, 34, … and then

use the rule to determine t15 of the sequence.

Solution

This sequence does not have a constant first difference. For this reason it is easier to put the sequence into a table like shown below. You may have noticed that it has a constant second difference.

As the second difference pattern is constant the rule must be of a quadratic =2 ++ =2 ++ nature. i.e. y ax bx c or tn an bn c .

=1 = The second difference pattern equals 2 then, a 2 (21) .

= When n = 0 we know that tcn .

Working backwards in the table we get c = -1.

=+−2 Substitute in a known point, like (1,2), we get, 21b ( 11) b = 2

=+−2 So, tnn 21 n.

We can work out t15 .

t =+−152 2 15 1 15 ( ) = 254

You can easily use your calculator to determine this rule. Make sure you can do so.

Finding the sequence from a rule

In the previous section we have found the rule for some sequences.

If a sequence is described by an algebraic expression it is easy to generate some of its terms.

=2 + ≥ For example, the sequence described by tnn 22 for n 1, will be

2 = += n = 1, t1 21( ) 2 4

2 = += n = 2, t2 2( 2) 2 10

2 = += n = 3, t3 2( 3) 2 20

2 = += n = 4, t4 2( 4) 2 34

So the sequence is 4, 10, 20, 34, …

Defining a sequence

We have now used three different methods to define a sequence, so it may be useful to restate these.

A sequence may be specified:

• by writing as many terms as required to establish a pattern

• by describing how to construct the list of terms, starting with the first and working down progressively. This requires a formula for calculating a particular term from the one that comes before it and is called a recurrence relation or recursive equation

• by writing a formula for the general term tn .

Note: The general term for a sequence is usually given the symbol tn , but this can

be lower case or upper case (capital letters), so Tn is equally acceptable.

Sometimes another letter is used to denote the general term. For example, =2 + in the sequence above we could use gn , i.e. gnn 22

Skills Development 3.1

1. For each of the sequences below, find an algebraic expression (or rule) to describe the sequence.

(a) 4, 8, 12, 16, 20, …

(b) 2, 3, 4, 5, …

(d) 2, 1, 0, -1, -2, …

2. Find the first 5 terms for the sequences generated by the following algebraic expressions.

= + ≥ (a) Tnn 25 for n 1

=1 2 + ≥ (b) Tnn 2 1 for n 1

= + ≥ (c) Tn nn( 1) for n 1

= n −1 ≥ (d) Tn 2 for n 1

24− 3n 3. (a) Determine the first five terms of the sequence T = . n 6

(b) Write a recursive definition to describe the sequence.

= − 4. A sequence has an algebraic expression Ynn 23 4 . Write a recursive definition that describes the sequence.

5. A taxi fare has a flagfall (fixed charge) of $3 and a 50 cents per kilometre charge. The fares form a sequence of costs defined by the nth term of n T =3 + , where T is the cost of a taxi journey of n kilometres. n 2 n

(a) Find the first three terms of the sequence (T1 , T2 , T3 ).

(b) Write the recursive definition for this sequence.

Lesson 4 Difference equations

By the end of this lesson you should be able to: • Write a difference equation for any sequence • See the connection between a difference equation and a recurrence equation • Find terms of a sequence from a difference equation.

Difference Equations

A difference equation highlights the change from one term to the next.

Consider the sequence 3, 8, 13, 18, …

We have previously established the recursive definition for this sequence as:

= + = ≥ ttnn+1 5 t1 3 for n 1

As this sequence has a constant first difference pattern (each term is 5 more than the term before it) it can be written as a difference equation.

− The difference between successive terms is given by ttnn+1 .

−= = ≥ For the sequence above ttnn+1 5 t1 3 for n 1

This equation is telling us that the difference between successive terms of the sequence is 5.

Algebraic approach

Using the recursive definition we have:

= + ttnn+1 5

If we rearrange the equation we get:

−= ttnn+1 5

This is the difference equation!

Example

Write down the first 5 terms of the sequence defined by

−=− = ≥ ttnn+1 2 t1 5 for n 1

Solution

We can think of this is two ways.

The difference between successive Change to recursive form by terms is -2. rearranging the difference equation. = If t1 5 then tt= − 2 =−= nn+1 t2 523 =−= t3 321 etc If n = 1

tt=−=23 etc The sequence is 5, 3, 1, -1, -3, … 21 The sequence is 5, 3, 1, -1, -3, …

Example

Write down the first 5 terms of the sequence defined by

−= = ≥ ttnnn+1 2 t1 5 for n 1

Solution

This difference equation is more difficult to interpret.

We can again think of this in two ways.

The difference between successive Change to recursive form by terms is 2n. rearranging the difference equation. −= If n = 1 tt2121( ) then = + tnn+1 tn2 tt= + 21 21 ( ) = + 52 If n = 1 = 7

= + = + tt3222( ) tt2121( ) If n = 2 =74 + etc =52 + etc = 11 = 7 The sequence is 5, 3, 1, -1, -3, … The sequence is 5, 7, 11, 17, … This approach shows that the difference equations are really the same as recursive equations.

Example

Determine the difference equation for the sequence 6, 10, 14, 18, …

Solution

This sequence has a constant first difference pattern (each term is 4 more than the term before it).

− The difference between consecutive term sis given by ttnn+1 .

∴ −= = ≥ ttnn+1 4 t1 6 for n 1

Note: You could also use a table or your calculator to find the difference equation.

Skills Development 4.1

1. Use tn and tn +1 to write a difference equation for each of the following sequences.

(a) 4, 6, 8, 10, …

(b) 20, 30, 40, …

2. Use tn and tn −1 to write a difference equation for each of the following sequences.

(a) 3, 6, 9, 12, …

(b) 10, 40, 70, …

3. For the following sequences

th (i) find tn , the rule for the n term, and

− (ii) write down the difference equation ttnn+1

(a) 1, 4, 9, 16, …

(b) 3, 5, 7, 9, …

(d) 3, 6, 12, 24, …

4. Use the difference equations to find the first 4 terms of each sequence.

−= = (a) ttnn+1 8 t1 3

−= = (b) ttnn+1 5 t1 8

−=− = (c) ttnn+1 2 t1 10

−= = (d) ttnn−1 7 t1 7

−=− = − (e) ttnn−1 6 t1 4

5. Write the first four terms of the sequences defined below.

−=2 = (a) tnn+1 tn t1 1

−= − = − (b) ttnnn+1 45 t1 5

n + 3 (c) tt−= t = 3 nn−1 4 2

6. A sequence is defined by the difference equation

−= = − TTnn+1 8 T1 15

(a) List the first four terms of the sequence.

(b) State the rule for the nth term of the sequence.

Lesson 5 Difference equations

By the end of this lesson you should be able to: • write a difference equation for any sequence • see the connection between a difference equation and a recurrence equation • find terms of a sequence from a difference equation.

Using difference equations

Let’s use difference equations for some practical applications.

Example

The successive rungs of a tapering ladder differ by a fixed amount. The difference between the lengths of the 5th and the 6th rungs is 4 cm.

If the first rung of the ladder is 56 cm, find a formula for the nth rung of the ladder.

Find the length of the 10th rung.

Solution

Since the lengths of the rungs are decreasing by 4cm, a difference equation −=− th can be written as tt65 4 where t6 = the length of the 6 rung.

−=− = The difference equation is: ttnn+1 4 and t1 56

−=− = We can also write this as: ttnn−1 4 and t1 56 .

This can be thought of as a recurrence relation as it can be rearranged to = − ttnn+1 4 (the next term is 4 less than the previous term).

To find an algebraic formula for the sequence, first write down some of the sequence.

= t1 56 = t2 52 = t3 48

This sequence has a constant difference of -4.

∴ =−+ tn 4 nc

To find the value for c, use any known term of the sequence.

t=−+4 nc n You could 56=−+ 4 1 c also place ( ) the terms in a c = 60 table to find the rule. ∴ =−+ tnn 4 60

∴ =−+= t10 4( 10) 60 20

The 10th rung is 20 cm long.

Example

A fisherman went fishing on 24 consecutive days and each Image

day caught 3 more fish than the previous day. The and his: catch. Fisherman Wikimedia. fisherman caught 10 fish on the first day.

Represent this situation using a difference equation, and show how to use it to determine the number of fish this fisherman caught on the 20th day.

(Note: A fisherman can be a man or a woman!)

Solution

The difference equation will be:

−= = ttnn+1 3 and t1 10

= t1 10 = += There is a difference of 3 between consecutive terms t2 10 3 13 = += t3 13 3 16

∴ the sequence is 10, 13, 16, …

We can easily determine a rule to find tn . If you can’t then….panic stations.

= + Tnn 37

To determine the number of fish this fisherman caught on the 20th day we have:

T =3 20 + 7 20 ( ) = 67

So the fisherman caught 67 fish on the 20th day.

Skills Development 5.1

1. Logs of wood are stacked in layers as shown in the diagram. There are 4 in the first row, 5 in the second row, 6 in the third, and so on.

(a) Write a recursive equation to show how the number of logs in each row is increasing.

(b) Find the number of logs in the nth row.

2. Tom has found by studying past costs that the model describing the cost of

producing his figurines is given by: Image: Smial,

−=− = CCfff+1 29 C1 31 Three figurines, Wikimedia. figurines, Three Where Cf is the cost (in dollars) to make f figurines.

(a) Find the cost of producing 2, 3, 4, 5, and 6 figurines.

(b) (i) Determine the number of figurines that need to be made for a minimum cost.

(ii) Determine the minimum cost.

The population of mosquitoes ( ) in my backyard depends on the number of litres of water used per week to water my garden. The difference equation is given by 𝑀𝑀 𝑥𝑥 𝑥𝑥 −=− = ≤≤ MMxx+1 66 2 x M1 166 for 0x 40

(a) Determine the number of mosquitoes in my garden when 2, 3, 4 and 5 litres of water are used per week.

(b) Show that the general rule Mx( ) =−++ x2 67 x 100 gives the difference equation above.

Lesson 6 Arithmetic sequences

By the end of this lesson you should be able to: • define an arithmetic sequence • find terms of an arithmetic sequence • see that the terms of any arithmetic sequence can be described by a general rule.

Suppose a business purchased office furniture worth $15000 at the beginning of 2010. For taxation purposes, this furniture is considered to depreciate in value by $900 each year.

At the beginning of 2011 the value will be $14100, and at the beginning of 2012 the value will be $13200, and so on.

The value of the furniture at the beginning of each year forms a linear sequence.

$15000, $14100, $13200, $12300, and so on …

These numbers can be thought of as a sequence for which the recursive equation is given by:

= − = ttnn−1 900 , t1 15000

Alternatively, we can say that the difference equation is:

−=− = ttnn−1 900 , t1 15000

= We can call t1 15000 because $15000 is the first term in the sequence, = t2 14100 because $14100 is the second term in the sequence and so on.

A sequence such as this (and many others you have been looking at in earlier lessons) shows that the difference between successive terms (the value of the furniture) is a constant. Such a sequence is called an arithmetic sequence.

The sequence on the previous page can also be called a linear sequence, because the graph of the terms of a sequence with ‘constant difference’ is a straight line.

Skills Development 6.1 1. Continue the sequence of furniture values (from earlier), so that you have written the first 10 terms.

$15000, $14100, $13200, $12300, ______, ______, ______, ______, ______, ______

2. Use the grid below to plot a graph of the value of the furniture against the = years. Assume t1 $15000 . The first few have been done for you, however you need to draw in the x-axis scale.

Since the value of the furniture can be considered to be decreasing evenly (i.e. it would be $13650 mid-way through 2011), then it is reasonable to join these points.

This graph shows why this method of depreciation is sometimes called ‘straight line depreciation’.

3. (a) What is the gradient of the graph?

−=− (b) How is the gradient related to the difference equation ttnn−1 900 ?

Activity

A ‘potato race’ is a novelty event often held on sports days. For this race a bucket is placed at one end and ‘potatoes’ are placed at regular intervals as shown in the diagram.

Image: Potatoes. Wikimedia.

The race starts and finishes at the bucket and each competitor must collect and put all of their potatoes into the bucket.

To retrieve the first potato, the competitor must run 12 m ( 6 m to the potato and 6 m back).

For the second potato, the distance to run is 18 m, (6 + 3 + 3 + 6), and so on.

Complete this table to show the distance run for each potato.

The distance run for each potato forms an arithmetic sequence because each ‘term’ is found by adding a constant number to the previous term.

= + ttnn−1 6

= + = The recursive equation ttnn−1 6 , t1 12 , is sometimes useful if we need to progress from term to term, but it becomes tedious if we want to know later terms,

such as t20 for instance.

We will now look at a way to find later terms, without lots of working out.

Just look at the ‘one-way’ distances for this potato race.

Use the symbol a to represent the distance between the bucket and the first potato and d to represent the difference between successive potatoes.

The sequence of (one-way) distances for this potato race is still an arithmetic sequence and can be shown to be:

t1 = a

t2 = a + d Do you see the pattern here? t3 = a + 2d

t4 = a + 3d etc

If there were n potatoes, then:

=+− tn an( 1) d

This formula can be used to describe any arithmetic sequence.

a = the first term

d = the difference between terms, called the common difference

and n = the number of terms

The sequence 3, 5, 7, 9, … is another example of an arithmetic sequence. We can see that the first term is 3, (i.e. a = 3) and the common difference is 2, (i.e. d = 2).

For this sequence:

=+− tn an( 1) d General formula

=+− =+− tnn 3( 12) 3 2 n 2

= + tnn 21 The ‘rule’

You can find the rule for any arithmetic sequence by substituting the values for a and d into the general rule.

Lesson 7 Arithmetic sequences

=+− The formula tn an( 1) dcan be used to describe any arithmetic sequence:

a = the first term

d = the difference between terms, called the common difference

and n = the number of terms

Example 1

For the sequence of furniture values (at the beginning of this lesson), i.e. 15000, 14100, 13200, 12300, … , complete the following.

(a) The first term, a is ______

(b) The common difference, d is ______

=+− tn an( 1) d

(d) Write the rule for this particular sequence ______

(e) What is the value of t10 (beginning of 2019) ______

Solution

(a) a = 15000

(b) d = -900

= +−− tnn 15000( 1) ( 900) (c) =15000 −+ 900n 900 = − tnn 14100 900

= − (d) tnn 14100 900

=−= (e) t10 14100 900( 10) 5100

Example 2

Find the 12th term and the 30th term for the arithmetic sequence

4, 10, 16, 22, …

Solution

For this arithmetic sequence we can see that a = 4 and d = 6.

The general rule is then

=+− tn an( 1) d

=+− =+− tnn 4( 16) ( ) 4 6 n 6

=−+ tnn 26

=−+ = To find t12 : t12 2 6( 12) 70

=−+ = To find t30 : t30 2 6( 30) 178

Example 3

= Finds the first four terms of an arithmetic sequence such that T5 7 and = T11 22 .

Solution

There are a few ways that you can think about this problem.

If we imagine the terms written in boxes like this, then we can put the numbers 7 and 22 in the 5th and 11th box.

There are 6 steps to get from 7 to 22, and since 22 – 7 = 15, then each step 15 must equal = 25. (i.e. d = 2.5). 6

There are 4 steps back to the first box, and 4×= 2. 5 10

∴ a =−=− 7 10 3

The first 4 terms are -3, -0.5, 2, 4.5

Look on the next page for an alternative solution.

Alternatively:

= = From T5 7 and T11 22 .

we get 7=+−ad( 51) and 22=+−ad( 11 1)

i.e. ad+=47 and 22=+−ad( 11 1)

Solving these equations simultaneously (easy to do on calculator), we find that a = −3 and d = 25. .

The first four terms are -3, -0.5, 2, 4.5

Skills Development 7.1

1. Which of the following form an arithmetic sequence?

For those sequences which are arithmetic, state the first term and the common difference.

(a) 3, 53, 103, 153, 203, …

(b) 8, 1, -6, -13, -20, …

1 1 1 1 (c) 1, 2 , 4 , 8 , 16 , …

(d) 1, 2, 5, 10, 20, 50, …

(e) 2++++ππππ, 22 , 23 , 24 , ...

(f) 2123,,,,,... 2 2 2 45 2

2. For the arithmetic sequences A and B which follow,

Sequence A -4, 0, 4, 8, 12, … Sequence B 20, 17, 14, 11, 8, …

(a) What is the first term, a?

(b) What is the common difference, d?

(d) Calculate the value for t20 for each sequence.

3. The first term and common difference are given for arithmetic sequences. Write down the first 4 terms and the 100th term of each.

(a) a = 5 , d = −3

(b) a = 18, d = 2

4. Find the rule for the nth term of the following arithmetic sequences and then determine the 60th term for each.

(a) 5, 8, 11, …

(b) -314, -302, -290, …

5. What are the first 4 terms of each of the arithmetic sequences for which the common difference and one term are given as:

= = − (a) T2 9 , d 3

= − = − (b) T5 1, d 2

6. Extending arms have been used on spacecraft because they are compact and streamlined. When fully extended the span of these arms depend on the number of ‘parallelograms’ used.

The span of the arms forms an arithmetic sequence.

(a) What is the first term, a, and the common difference, d, for this sequence?

th (b) Write a rule for Tn , the number of metres in the span for the n arm.

(d) If the span is 8.6 metres, how many parallelograms are in the span?

7. If a sequence started at -16 and could only add multiples of 7, could you reach 685 exactly? If not, how close can you get to 685? How many sevens have been added?

8. The nth term of an arithmetic sequence is (6 4). Which of the following belong to this sequence? 1 4 𝑛𝑛 − (a) 6 (b) 7 (c) 8 (d) 9 (e) 10?

Lesson 8 Geometric sequences

By the end of this lesson you should be able to: • Recognise sequences whose differences are not constant • Define a geometric sequence.

Sequences whose differences are not constant

The sequence 2, 4, 8, 16, … has differences that are not the same. We can still describe the sequence as:

= = (i) ttnn2 −1 , t1 2

(ii) In words, such as: “The first term is 2 and each term is twice the previous term”.

= n 1 2 3 (iii) tn 2 (See how the sequence is 2 , 2 , 2 , and so on ...)

Note: Compare the last description to the equation y = 2x . Do you see the = n x similarity of tn 2 and y = 2 ?

x = n Here is the graph of both y = 2 and the sequence tn 2 .

= n The sequence described by tn 2 represents points on the exponential curve y = 2x .

Geometric sequences

When each term of a sequence is formed by multiplying the preceding term by a constant, the sequence is called a geometric sequence.

This means that a geometric sequence is a special kind of sequence in which:

t= tr × nn+1

t n +1 = r tn

t r is called the common ratio because n +1 , the ratio between any two consecutive tn terms, is always constant.

The sequence 3, 6, 12, 24, 48, … is a geometric sequence because each term is multiplied by the same value to get the next term. For this sequence the common ratio is 2.

The sequence 100, 50, 25, 12.5, … is a geometric sequence because the constant multiplier is 0.5. The common ratio is 0.5.

A growth pattern

A scientist is studying the growth of a surface weed in a specially treated lake. He found that under the prevailing conditions the area covered by the surface weed is increasing by 20% every day.

On the first day the area covered was 100 cm2 , on the second day the area was 120 cm2 , and so on.

Complete this table to indicate the expected area covered by the weed in the lake with t, the number of days. (Give area to the nearest cm2 .)

Time 1 2 3 4 5 6 7 8 9 10 (days)

Area 100 120 144 173 covered ( cm2 )

Did you increase the quantity by 20%, or multiply by a factor of 1.20?

Use the results from the table to complete the graph of the area covered by the weed over a 10 day period.

Use the graph to estimate the area of lake covered by weed after 12 days.

Your estimate: ______

From the table on the previous page determine a rule to calculate the area of weed in the lake after 12 days.

Your rule: ______

Area from rule: ______

Compare your estimates with the solutions at the back of this booklet.

If we think of the numbers from the table as numbers of a sequence, then clearly they do not form an arithmetic sequence, because there is not a constant difference.

If you think about how the area covered was calculated, you should notice that each of these numbers form a sequence where the first term is 100 and each term is 1.20 times the previous term.

The table below shows how the area for the previous problem has been calculated.

You can work the rest out here.

The values for ‘area’ represent terms of a geometric sequence because each term is found by multiplying the previous term by the same number.

If we use the symbol a for the first term, and the symbol r for the common ratio, the terms of the sequence would be as given in brackets below.

= = A1 100 ( ta1 )

= × = A2 100 1. 20 ( t2 ar )

2 = × = 2 A3 100( 1. 20) ( t3 ar ) etc …

n −1 = × = n −1 and An 100( 1. 20) ( tn ar )

Sometimes the first term may be called A0 because it corresponds to ‘zero’ time

having elapsed during the measurements, however the general symbol is t1 because it is the first term.

If you find the first term referred to as t0 , then

= ta0

= t1 ar

= 2 t2 ar and so on

= n tn ar

What we really mean here is that n values start from 0, i.e. n = 0, 1, 2, …

Lesson 9 Geometric sequences

By the end of this lesson you should be able to: • Recognise sequences whose differences are not constant • Define a geometric sequence.

General formula for a geometric sequence

For any geometric sequence whose first term is a and common ratio is r,

= n −1 tn ar for n = 1, 2, 3, 4, …

= n tn ar for n = 0, 1, 2, 3, …

The recurrence relation for both of these is the same, that is:

t = × n = ttnn−1 r or r tn −1

Example

(a) Show that the sequence 2, 5, 12.5, 31.25, 78.125, … is a geometric sequence.

(b) What is the value of the first term, a, and the common ratio r?

(d) What is the recursive definition for this sequence?

Solution

(a) Look at the terms of the sequence. We can check to see if there is a common multiplier by finding the ratio of successive terms.

T 5 T 12. 5 T 31. 25 2 = = 25. 3 = = 25. 4 = = 25. T1 2 T2 5 T3 12. 5

You can convince yourself that this pattern will continue.

Since each term is found from the previous one by multiplying by a constant value, the sequence is a geometric one.

(b) First term, a = 2, and the common ratio, r = 2.5.

= n −1 (c) For a geometric sequence, tn ar .

= × n −1 So, tn 2 25. .

(d) A recursive definition is one which describes the sequence by referring = = back to the previous term. For this sequence, ttnn+1 25. , and t1 2 .

Example

For a particular geometric sequence the 5th term is 2.56 and the 7th term is 40.96. Find the common ratio and the first and second terms for this sequence.

Solution

There are two ways to think through this problem.

If we imagine the terms written in boxes like this, then we can put the numbers 2.56 in the 5th box and 40.96 in the 7th box.

There are 2 steps to get from 2.56 to 40.96, and we need to multiply by the common ratio at each step.

= × We can use the general form of a geometric sequence, ttnn−1 r to help.

This shows that:

2.. 56××rr = 40 96 40. 96 r 2 = = 16 2. 56

If r 2 = 16 , then r =±=±16 4 . There are 2 possible values for r.

Alternatively

= = t5 2. 56 t7 40. 96 =ar × 51− and =ar × 71− 2. 56 =ar × 4 40. 96 =ar × 6

By division:

40. 96 ar× 6 = 2. 56 ar× 4 16 = r 2 r = ±4

Now we need to find the values of the first term, a, and the second term for each of the values of r.

(51− ) 4 (i) If r = 4, 2. 56 =×=×ar a(4)

2. 56 So a = = 0. 01 256

t=×= ar 0. 01 × 4 Second term: 2 = t2 0. 04

(51− ) 4 (ii) If r = -4, 2. 56 =ar × = a ×−( 4)

2. 56 So a = = 0. 01 256

t= ar × =0. 01 ×−( 4) Second term: 2 = − t2 0. 04

Skills Development 9.1

1. Find the first four terms for the following sequences. (a) The first term is 40 and the common ratio is 1.5.

(b) The first term is 200 and the common ratio is -2.

2. For the following geometric sequences, determine the first term, a, and the

common ratio, r, and the rule for tn .

(a) 1.5, 6, 24, 96, …

(b) 1, -2, 4, -8, 16, …

3. The first two terms of a geometric sequence are 500 and 100.

(a) What is the third term?

(b) Give an expression for tn in terms of a, r and n.

4. What is the common ratio and the first term for the geometric sequence for which:

= = (a) T2 10 and T5 1250

= = (b) t3 120 and t5 76. 8

Lesson 10 Application of sequences

By the end of this lesson you should be able to: • Apply understanding of arithmetic and geometric sequences to real world situations. • Apply understanding of Fibonacci sequences to real world situations. • Test conjectures systematically by checking cases and searching for counter examples.

Applications of sequences

We can use sequences to describe and solve real situations in our world.

Problems involving arithmetic sequences (linear growth or decline) are found in areas such as:

• number theory (property of numbers)

• combinatorics (study of countable things)

• computer science (study of information and computation).

Image: J Williams, Mazatlan diver sequence. Wikimedia.

Example

The auditorium of a small theatre has 30 seats in the first row, 34 in the second row, 38 in the third row and so on. There are 25 rows all together.

Image: C Culler, Fletcher Hall, Wikimedia. How many people can fit into this theatre?

Solution

The number of seats in each of the rows forms an arithmetic sequence, with the first term a = 30 and the common difference d = 4.

You should be = + = The recursive rule can be determined to be TTnn+1 4 , T1 30 . able to determine these. Refer to earlier The general term can be determined to be Tn=4 + 26 . lessons for n support. There are 30 seats in the first row and 126 seats in the 25th row.

We can add the seats in each row to get 1950 seats in total.

Problems involving geometric sequences (exponential growth or decay) are found in areas such as:

• geometry (study of space)

• number theory (property of numbers)

• physics (study of matter)

• engineering, biology, economics, queuing theory, finance, etc.

Example

The world population of a certain species of bird was 5000 at the beginning of 1999, and is decreasing such that the population decreases by 10% each year. In which year will the bird population drop below 1000?

Image: W Siegmund, Chen caerulescens, Wikimedia. Solution

Note that a decrease of 10% means we multiply by a factor of 0.90.

We can represent this situation in a table. You can finish it.

This table shows that the ratio of successive terms is 0.90, so the bird population can be modelled by the terms of a geometric sequence. It is graphically shown below.

We wouldn’t expect the bird population to be 5000 for all of 1990 and then on the 31st of December to suddenly drop to 4500. the population would be decreasing evenly over the year, so a graph of this data would show the points joined by a curve.

The graph shows that after 9 years the bird population is just below 2000.

We can use what we know about geometric sequences to find during which year the bird population will first drop below 1000.

The general rule for the bird population is

= × n Pn 5000 0. 9

th Where Pn is the population at the end of the n year.

= We need to find the value of n when Pn 1000 .

1000= 5000 × 0. 9n

Use your calculators SOLVER feature to get n = 15. 27 .

To interpret this, we can draw a time line like the one below. This shows that n = 15. 27 is during the year 2005.

Skills Development 10.1

1. Joe starts a new job on 1st January 2010 and during 2010 will receive a salary of $72000. His contract guarantees a 5% salary increase on each subsequent 1st January until and including 1st January 2015. Calculate Joe’s expected salary for each year from 2011 to 2015.

2. The predicted population growth for a town that now has 2000 people, is for an increase of 300 in the first year, 360 in the second year, 432 in the third year, and so on.

Given that the pattern continues, find

(a) The approximate increase in the 7th and 8th years.

(b) When the population increase is ~1290.

3. In a certain country the number of people over 65 years of age is increasing by 2.5% each year. In 2007 the number of ‘over 65’s’ was 1.8 million.

(a) How many people will be in the over 65 age group after

(i) one year

(ii) two years

(iii) 10 years

(b) If this rate of increase continues, during which year will the number of ‘over 65’s’ first be greater than 2.5 million?

4. Ron owes the Taxation Office for unpaid taxes of $27950. He set up a repayment schedule whereby automatic deductions of $200 a fortnight occur from his savings account. Find a linear rule and show how you can use it to determine how many weeks it will take Ron to pay off the debt.

5. (Not on syllabus – but an interesting task): One mile is about 1.609 kilometres. A very good approximation to convert to and from kilometres and miles is to look at consecutive numbers in the Fibonacci sequence. For example, 5 miles is converted to be 8 kilometres, 8 miles is converted to be 13 kilometres, and so on. Test that this conversion works for other Fibonacci numbers and compare each conversion with the real conversion, using 1 mile = 1.609 km.

6. The geometric progression 1, 2, 4, 8, 16, 32, … is important in number theory. If the sum of the first n terms of this progression is a prime number, then this sum, times the nth term, is a perfect number. A perfect number is a number that is half the sum of the positive divisors (factors), including itself. The sequence is: 6, 28, 496, 8128, … For example, the sum of the first 3 terms of the series 1 + 2 + 4 = 7, this is a prime number. The sum 7 multiplied by 4 (the 3rd term) equals 28, which is a perfect number. Test this conjecture works for the next perfect number.

7. Every positive integer can be uniquely represented as a sum of different ‘powers of two’. Test this generalisation.

Lesson 11 Examination question practice – Calculator Free

By the end of this lesson you should be able to: • practise your learning using Calculator Free examples • apply understanding of arithmetic and geometric sequences to exam type questions.

Examination Type Questions – Calculator Free

Using 2019 ATAR Maths Applications course examination (School Curriculum and Standards Authority).

Check solutions at the end after you have had a go at these questions.

Formulae Sheet (relevant formulae)

Question 6 2019 (6 marks)

The population of turtles in an artificial lake at a wildlife sanctuary is initially 32 and research has shown a natural decrease in population of 50% each year. Twenty extra turtles are introduced to the lake at the end of each year.

(a) Determine a recursive rule for the turtle population. (2 marks)

(b) Determine the long-term steady state of the turtle population.(2 marks)

(c) If the wildlife sanctuary preferred a long-term steady state of 80 turtles, what yearly addition of turtles would be required to produce this steady state? Assume all other conditions remain the same.

(2 marks)

A researcher compared the performance of various golf balls. The graph below shows the height reached above the ground by a particular golf ball after each of the first three bounces. It was initially dropped from a height of 54 cm.

(a) Write the recursive rule for this sequence. (3 marks)

th (b) Write the rule for the n term of this sequence. (1 mark)

(2 marks)

Lesson 12 Examination question practice – Calculator Assumed

By the end of this lesson you should be able to: • practise Calculator Assumed Questions • apply understanding of arithmetic and geometric sequences to exam type questions.

Examination Type Questions – Calculator Assumed

Using 2019 ATAR Maths Applications course examination (School Curriculum and Standards Authority).

Check solutions at the end after you have had a go at these questions.

Relevant formulae

Question 7 2019 (6 marks)

A water tank is full. When a tap at the bottom of the tank is opened, 84 litres run out in the first minute, 78 litres in the second minute and 72 litres in the third minute. This pattern continues until the tank is empty.

(a) Write a rule for the nth term of a sequence in the form = + , which will model this situation where is the amount of water that runs out in the nth 𝑛𝑛 minute. 𝑇𝑇 𝐴𝐴 (2𝐵𝐵 marks)𝐵𝐵 𝑇𝑇𝑛𝑛

(b) How many litres run out in the seventh minute? (1 mark)

(d) What is the capacity of the tank? (2 marks)

Question 10 (7 marks)

Ruby Ducks Coffee shops commenced operations in 1992 and had 15 stores open by the end of the year. They have been so successful over the years that the number of stores worldwide has continued to grow exponentially since then. The number of shops operating, , at the end of 2017 was 22 579 and at the end of 2018 was 30 256. 𝑇𝑇 The number of shops operating at the end of n years can be represented by the recursive rule = 1.34 – 1, = 15.

(a) Show mathematically𝑇𝑇𝑛𝑛 𝑇𝑇 that𝑛𝑛 the𝑇𝑇1 common ratio is approximately 1.34. (1 mark)

(b) Write the rule for the nth term of this sequence. (1 mark)

Typically, each store has twelve employees working during the day across different shifts. Each employee earns, on average, $114.80 per day.

(d) Calculate the total daily wages for all stores at the beginning of 2012.

(3 marks)

Further Practice Examination question practice

By doing further practice you should be able to: • apply understanding of arithmetic and geometric sequences to exam type questions.

Examination Type Questions

https://www.scsa.wa.edu.au/ or contact your school.

Other video resources: Khan Academy https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:sequences

Glossary

Growth and decay in sequences Arithmetic sequence An arithmetic sequence is a sequence of numbers such that the difference between any two successive members of the sequence is constant. For example, the sequence 2, 5, 8, 11, 14, 17, … is an arithmetic sequence with first term 2 and common difference 3. By inspection of the sequence, the rule for the term of this sequence is: 𝑡𝑡ℎ 𝑛𝑛 𝑡𝑡𝑛𝑛 = 2 + ( 1)3 = 3 1 1 If is used to denote the term in the sequence, then a recursion 𝑡𝑡𝑛𝑛 𝑛𝑛 − 𝑛𝑛 − 𝑛𝑛 ≥ relation that will generate this𝑡𝑡ℎ sequence is: = 2, = + 𝑡𝑡𝑛𝑛 𝑛𝑛 3 1 𝑡𝑡1 𝑡𝑡𝑛𝑛+1 𝑡𝑡𝑛𝑛 𝑛𝑛 ≥

First-order linear A first-order linear recurrence relation is defined by the rule: recurrence relation = , = + for 1

1 𝑛𝑛+1 𝑛𝑛 = 10, = 5 + 1 1 𝑡𝑡For example,𝑎𝑎 𝑡𝑡 the𝑏𝑏 𝑏𝑏rule: 𝑐𝑐 𝑛𝑛 ≥ for is a first- order recurrence relation. 𝑡𝑡1 𝑡𝑡𝑛𝑛 𝑡𝑡𝑛𝑛−1 𝑛𝑛 ≥ The sequence generated by this rule is: 10, 51, 256, … as shown below.

= 10, = 5 + 1 = 5 × 10 + 1 = 51, = 5 + 1 = 5 × 51 + 1 = 256, … 𝑡𝑡1 𝑡𝑡2 𝑡𝑡1 𝑡𝑡3 𝑡𝑡2 Examples of the use of first-order linear recurrence relations applied to practical problems include; investigating the growth of a trout population in a lake recorded at the end of each year and where limited recreational fishing is permitted, or the amount owing on a reducing balance loan after each payment is made.

Geometric growth or A sequence displays geometric growth or decay when each term is decay (sequence) some constant multiple (greater or less than one) of the preceding term. A multiple greater than one corresponds to growth. A multiple less than one corresponds to decay. For example, the sequence: 1, 2, 4, … displays geometric growth because each term is double the previous term. 100, 10,1, 0.1, … displays geometric decay because each term is one tenth of the previous term. Geometric growth is an example of exponential growth in discrete situations. Examples of practical situations where such sequences could be applied include: analysing a compound interest loan or investment; the growth of a bacterial population that doubles in size each hour; the decreasing height of the bounce of a ball at each bounce; or calculating the value of office furniture at the end of each year using the declining (reducing) balance method to depreciate.

Geometric sequence A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. For example, the sequence 2, 6, 18, ... is a geometric sequence with first term 2 and common ratio 3. By inspection of the sequence, the rule for the term of this sequence is: 𝑡𝑡ℎ 𝑛𝑛 = 2 × 3 1 𝑛𝑛−1 𝑡𝑡If𝑛𝑛 is used to denote𝑛𝑛 ≥ the term in the sequence, then a recursion relation that will generate this𝑡𝑡ℎ sequence is: = 2, = 3 𝑡𝑡𝑛𝑛 𝑛𝑛 1 𝑡𝑡1 𝑡𝑡𝑛𝑛+1 𝑡𝑡𝑛𝑛 𝑛𝑛 ≥ Linear growth or decay A sequence displays linear growth or decay when the difference (sequence) between successive terms is constant. A positive constant difference

corresponds to linear growth, while a negative constant difference corresponds to decay. Examples: The sequence, 1, 4, 7, … displays linear growth because the difference between successive terms is 3. The sequence, 100, 90, 80, … displays linear decay because the difference between successive terms is –10. By definition, arithmetic sequences display linear growth or decay. Examples of practical situations where such sequences could be applied include: analysing a simple interest loan or investment; calculating a taxi fare based on the flag fall and the charge per kilometre; or calculating the value of an office photocopier at the end of each year using the straight-line method or the unit cost method of depreciation.

Recursion See Recurrence relation.

Recurrence relation A recurrence relation is an equation that recursively defines a sequence; that is, once one or more initial terms are given, each further term of the sequence is defined as a function of the preceding terms.

Sequence A sequence is an ordered list of numbers (or objects). For example, 1, 3, 5, 7 is a sequence of numbers that differs from the sequence 3, 1, 7, 5 because order matters. A sequence maybe finite, for example, 1, 3, 5, 7 (the sequence of the first four odd numbers), or infinite, for example, 1, 3, 5, … (the sequence of all odd numbers). Copyright © School Curriculum and Standards Authority 2019

Summary

A number sequence is a set of numbers arranged in a definite order. Each number of the sequence is called a term.

The general term for a sequence is usually given the symbol tn , but this can be either

lower case or upper case (capital letters), so Tn is equally acceptable. The subscript, n, tells us the term number. The rule which tells you how to get from one term to the next is called a recurrence relation. A difference equation highlights the change from one term to the next. A sequence such that the difference between successive terms is a constant is called an arithmetic sequence. =+− tn an( 1) d This formula can be used to describe any arithmetic sequence. a = the first term

d = the difference between terms, called the common difference and n = the number of terms When each term of a sequence is formed by multiplying the preceding term by a constant, the sequence is called a geometric sequence. = × tnn+1 tr or t n +1 = r tn t r is called the common ratio because n +1 , the ratio between any two consecutive tn terms, is always constant.

Solutions

Lesson 1

SD 1.1

1. (a) 13 (b) 3

2. (a) 6 (b) 0

SD 1.2

1 (i) (a) 25, 29

(b) add 4 each time

tn +1 = 21

(ii) (a) 6.25, 3.125

(b) divide by 2

tn +1 = 12.5

(iii) (a) -5, -8

(b) subtract 3

tn +1 = 1

(iv) (a) 35, 48

(b) amount being added increases by 2 each time

tn −1 = 8

(v) (a) 64, 128

(b) multiply by 2

tn +1 = 128

(vi) (a) 33, 65

(b) amount being added doubles each time

tn −1 = 5

Lesson 2

SD 2.1

= = 1. (a) (i) ttnn4 −1 , t1 2

= = (ii) ttnn+1 4 , t1 2

= 1 = (b) (i) ttnn2 −1, t1 100

= 1 = (ii) ttnn+1 2 , t1 100

= − = (c) (i) ttnn−1 4 , t1 20

= − = (ii) ttnn+1 4 , t1 20

= + = (d) (i) ttnn−1 15. , t1 4

= + = (ii) ttnn+1 15. , t1 4

2. (a) 3, 9, 27, 81, 243 1 g =1 + (b) 2 and so on … g1 = 2

2 3 1, 2, 1.5, 1 3 , 1 5

2 hh= (c) 21( ) and so on … = 4 2, 4, 16, 256, 65536

64 t = (d) 2 and so on … t1 = 4 4, 4, 4, 4, 4

tt= +×32 (e) 21 and so on … = 10

4, 10, 19, 31, 46

tt= + 3 t (f) 32 1 and so on … = 5

1, 2, 5, 11, 26

3. (a) = 0.95 + 12, = 200

(b) 𝐶𝐶𝑛𝑛= 207 𝐶𝐶𝑛𝑛−1 𝐶𝐶0 (c) = 220, therefore the 14th month 4 𝐶𝐶 14 4. (a)𝐶𝐶 = 1.04 = 1500 (b) 𝑃𝑃𝑛𝑛 𝑃𝑃𝑛𝑛−1 𝑃𝑃0

7 = 1973, 8 = 2052, i.e. during the eight month.

𝑃𝑃(c) = 1.04𝑃𝑃 1 400, 0 = 1898 (Use the value for 6 as the starting value for the new relation). 𝑇𝑇𝑛𝑛 𝑇𝑇𝑛𝑛− − 𝑇𝑇 𝑃𝑃

(d) Using this model, the population of crickets would drop to zero in the 6th month (e) 75 crickets (4% of 1898 is 75.92)

(a) = (0.82) + 8 , = 430

𝑉𝑉𝑛𝑛 𝑉𝑉𝑛𝑛−1 𝑉𝑉0 (b)

4 = 219 to the nearest whole number (c) = 109, = 97, therefore, after 10 months 𝑉𝑉 𝑉𝑉9 𝑉𝑉10

Lesson 3

SD 3.1

1. (a) n 0 1 2 3 4

0 4 8 12 16

Constant 1st dp = Tnn 4

= + (b) Tnn 1

=2 − (c) Tnn 1

=−+ (d) Tnn 3

= + ≥ 2. (a) Tnn 25, n 1

7, 9, 11, 13, 15.

=1 2 + ≥ (b) Tnn 2 1, n 1

1.5, 3, 5.5, 9, 13.5.

= + ≥ (c) Tn nn( 1) , n 1

2, 6, 12, 20, 30.

= n −1 ≥ (d) Tn 2 , n 1

1, 2, 4, 8, 16.

3. (a) 3.5, 3, 2.5, 2, 1.5.

= − = (b) TTnn+1 05. , T1 35.

= − = TTnn−1 05. , T1 35.

= − 4. Tnn 23 4

T =23 − 4 T =23 − 8 1 2 = 19 = 15

T =23 − 12 T =23 − 16 3 4 etc = 11 = 7

∴ 19, 15, 11, 7, …

∴ = − = TTnn+1 4 , T1 19

= − = TTnn−1 4 , T1 19 .

5. (a) 3.5, 4, 4.5.

= + = (b) TTnn−1 05. , T1 35.

= + = TTnn+1 05. , T1 35.

Lesson 4

Full working out is not usually shown in these solutions. However you must always show full working out to gain full marks.

SD 4.1

−= = 1. (a) ttnn+1 2 t1 4

−= = (b) ttnn+1 10 t1 20

−=− = (c) ttnn+1 5 t1 50

−= = 2. (a) ttnn−1 3 t1 3

−= = (b) ttnn−1 30 t1 10

−=− = (c) ttnn−1 20 t1 80

= 2 3. (a) (i) Tnn

2 = + (ii) Tnn +1 ( 1)

2 −=+ −2 TTnnn+1 ( 1) n

−= + TTnnn+1 21 = T1 1

= + (b) (i) Tnn 21

−= (ii) TTnn+1 2 = T1 3

= + (c) (i) Tnn 41

−= (ii) TTnn+1 4 = T1 5

(d) (i) Exponential. Some effort is required to find the rule.

n = Tn 15. ( 2)

n +1 = (ii) Tn +1 15. ( 2)

nn+1 −= − TTnn+1 15..( 2) 15( 2) Simplified to: n −= TTnn+1 15. ( 2) = T1 3

= + 4. (a) ttnn+1 8

3, 11, 19, 27.

= + (b) ttnn+1 5

8, 13, 18, 23.

= − (c) ttnn+1 2

10, 8, 6, 4.

= + (d) ttnn−1 7

7, 14, 21, 28.

= − (e) ttnn−1 6

-4, -10, -16, -22.

= + 2 5. (a) tnn+1 tn

1, 2, 6, 15.

=+− (b) tnn+1 tn45

-5, -6, -3, 4.

n + 3 (c) tt= + nn−1 4

1 3 3, 4, 5 4 , 6 4 .

6. (a) TT= + 8 n+1 n -15, -7, 1, 9. (b) Using a table or your calculator you can get: = − Tnn 8 23

= −= (c) T20 8( 20) 23 137

Lesson 5

SD 5.1

(Handwritten.)

Lesson 6

SD 6.1

1. 11400, 10500, 9600, 8700, 7800, 6900

2. Put in an appropriate x-scale and plot the values. Then join as a straight line.

3. (a) m = −900

(b) Gradient is the difference.

Lesson 7

SD 7.1

1. (a) arithmetic, a = 3, d = 50

(b) arithmetic, a = 8, d = -7

(d) not arithmetic

(e) arithmetic, a = 2 + π , d = π

(f) not arithmetic

2. Sequence A:

(a) a = −4

(b) d = 4

tn=−+4 − 14 (c) n ( ) =48n −

= −= (d) t20 4( 20) 8 72

Sequence B:

(a) a = 20

(b) d = −3

tn=20 +−− 1 3 (c) n ( ) ( ) =23 − 3n

=−=− (d) t20 23 3( 20) 37

3. (a) 5, 2, -1, -4

=+− tn an( 1) d =+−−5(n 13) ( ) = − tnn 83

=−=− t100 8 3( 100) 292

(b) 18, 20, 22, 24

=+− tn an( 1) d =+−18(n 1) ( 2) = + tnn 16 2

=+= t100 16 2( 100) 216

=+− tn an( 1) d =−+−20(n 1) ( − 5) =−− tnn 15 5

=−− =− t100 15 5( 100) 515

4. (a) a = 5 , d = 3

tn=+−5 13 n ( ) =32n +

= += t60 3( 60) 2 182

(b) a = −314 , d = 12

tn=−+−314 1 12 n ( ) ( ) =12n − 326

= −= t60 12( 60) 326 394

= = − 5. (a) T2 9 and d 3

= T1 12 = = T2 9 T4 12 = T3 6

= − = − (b) T5 1 and d 2

T = −1 5 T = 5 T = 1 2 4 T = 7 = 1 T3 3

6. (a) a = 08. and d = 06.

Tn=08.. +− 1 06 (b) n ( ) ( ) =06..n + 02

= += (c) Tm10 0610.( ) 02 .. 62

06.n += 02 .. 86 (d) 06..n = 84 n = 14

∴ 14 parallelograms

7. -16, -9, -2, 5, 12, …

= 16 + ( 1)(7)

= 7 23 𝑡𝑡𝑛𝑛 − 𝑛𝑛 − Does 7 𝑛𝑛23− = 685?

No, it equals𝑛𝑛 − 101.14.

So cannot reach 685 exactly.

Closest is 684.

7 23 = 684

= 101 𝑛𝑛 − 𝑛𝑛= 684

𝑡𝑡So101 101 7’s have been added.

8. = (6 4) 1 𝑇𝑇𝑛𝑛 4 𝑛𝑛 − (a) (6 4) = 6 1 64 𝑛𝑛 4−= 24 = 28/6 𝑛𝑛 − Not𝑛𝑛 in sequence.

(b) (6 4) = 7 1 64 𝑛𝑛 4−= 28 = 32/6 𝑛𝑛 − 7 𝑛𝑛is not a term of the sequence.

= 6 𝑛𝑛 − 8 is𝑛𝑛 in sequence.

(d) (6 4) = 9 1 64 𝑛𝑛 4−= 36 = 40/6 𝑛𝑛 − 𝑛𝑛

Not in sequence.

(e) (6 4) = 10 1 64 𝑛𝑛 4−= 40 = 44/6 𝑛𝑛 − Not𝑛𝑛 in sequence.

Lesson 8

Time 1 2 3 4 5 6 7 8 9 10

Area 100 120 144 173 207 249 299 358 430 516

Approximately 800 sqm

11 = ≈ T12 100( 1. 20) 743

Lesson 9

SD 9.1

1. (a) 40, 50, 90, 135.

(b) 200, -400, 800, -1600.

= × n −1 2. (a) a = 1.5, r = 4, tn 15. 4

n −1 = − (b) a = 1, r = -2, tn ( 2)

= 3. (a) t3 20 n −1 (b) = 1 tn 500 ( 5 ) = 1 = (c) ttnn+1 5 , t1 500

= ∴= 4. (a) T2 10 ar 10

= ∴=4 T5 1250 ar 1250

ar 4 1250 = ar 10 r 3 = 125 r = 5

a = 2 and r = 5

= ∴=2 (b) t3 120 ar 120 = ∴=4 t5 76. 8 ar 76. 8

ar 4 76. 8 = ar 2 120 r 2 = 0. 64 r = ±08.

a =120 = 187. 5 , r = 0.8 08. 2

a =120 = 187. 5 , r = -0.8 −08. 2

Lesson 10

SD 10.1

1. a = 72000 , r = 1. 05

2011 72000×= 1.$ 05 75600

2012 72000×= 1.$ 052 79380

2013 72000×= 1.$ 053 83349

2014 72000×= 1. 054 $. 87516 45

2015 72000×= 1. 055 $. 91892 27

2. (a) 300, 360, 432, …

a = 300 and r = 12.

= 6 T7 ar 6 = 300( 1. 2) = 896

= 7 T8 ar 7 = 300( 1. 2) = 1075

= n −1 (b) Tn ar n −1 1290= 300( 1. 2)

Using SOLVER

n = 9 years.

3. (a) (i) 1.. 8×= 1 025 1 . 845

So ~1.845 million

(ii) ~1.89 million

(iii) ~2.30 million

= n −1 (b) Tn ar n −1 = 25. 181025 ..( ) n = 14. 3

So during 2021.

4. Arithmetic sequence.

a = 27950, d = -200

=+− Dn an( 1) d Dn=27950 +− 200 200 n =27750 − 200n

Find n for when Dn equals 0.

0= 27750 − 200n 200n = 27750 n = 138. 75

Note that n is the number of fortnights. So Ron will pay off his debt in 277.5 weeks.

5. Testing: 13 = . 8 mile = 13 km 8 1 625

21 = . 13 mile = 21 km 13 1 615

34 = . 21 mile = 34 km 21 1 619 and so on…

As the conversions use larger and larger Fibonacci numbers, the conversion between miles/kilometres gets more accurate.

6. Some trial & error is needed.

1 + 2 + 4 + 8 = 15 (not prime)

1 + 2 + 4 + 8 + 16 = 31 (prime!)

So the sum of first 5 terms is 31, a prime. The 5th term is 16. So we get the next perfect number, 31×= 16 496 .

7. You would firstly list the first few of the sequence of ‘powers of two’.

22220123, , , ,...

I can only use the following numbers in my sums!

1, 2, 4, 8, 16, …

Then trial some more sums:

5 = 1 + 4

6 = 4 + 2

7 = 4 + 2 + 1

And so on…

Lesson 11

Qn 6

Lesson 12

Qn 7 2019

Qn 10

End of Solutions