JC Success - Maths 2 - Website Sample

Published by 4schools

89F Lagan Road, Dublin Industrial Estate, Glasnevin, Dublin 11, D11 F98N, Republic of Ireland. T: ++ 353 1 8081494 - F: ++ 353 1 836 2739 - E: info@4schools.ie © 2022

ISBN 978-1-907330-50-6

All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, including photocopying and recording, without the publisher’s written permission. Such written permission must also be obtained before any part of this publication is stored in a retrieval system of any nature. Permission requests should be directed to 4schools, 89F Lagan Road, Dublin Industrial Estate, Glasnevin, Dublin 11, D11 F98N, Republic of Ireland. or info@4schools.ie 4schools has attempted to contact all proprietors of images, texts and graphics included in this book. Proprietors we could not reach are requested to contact us.

E-BOOK How to claim and access your e-Book Code:

To claim your e-Book

1. Log in to 4schools.examcraftgroup.ie/user/login

2 Activate your e-Book using the code above at 4schools.examcraftgroup.ie/activatecode

3. Download the app at 4schools.examcraftgroup.ie/apps

4. Log in to the app (please use the same login and password you used on the 4Schools website) 5. Download the e-Book

Notes

• Please ensure the email address used to complete the steps above is the email address of the student or teacher who will be using this e-Book.

• If the e-Book code was purchased using another email address, e.g., a parent’s, please register the email address of the e-Book user before completing the steps above. You can register the alternative email address here: 4schools.examcraftgroup.ie/user/register

Online Resources

Get access to exclusive content by scanning the QR code. https://4schools.examcraftgroup.ie/premium-content/junior-cycle-success-maths-book-2

Steps to Success SAMPLE

Introduction 03

Maths Exam Language 09

Topic 1 | Working with Sets 11

Topic 2 | Working with Statistics 33

Topic 3 | Working with Probability 53

Topic 4 | Working with Patterns 75

Topic 5 | Working with Functions & Graphs 99

Topic 6 | Working with Word Problems 131

Sample Papers 147

Throughout the book, QR codes are used. Once scanned, these will allow you access to our bank of online resources to further assist you in your exam preparation.

Meet the authors

Brendan O’Sullivan

Brendan O’Sullivan teaches at Davis College, Mallow, Co. Cork. He is an experienced examiner, creator of mock papers and designer of professional development. A regular contributor at educational conferences, he has a special interest in mathematics textbooks, having conducted one of the largest analyses of textbook tasks nationally and internationally as part of his PhD at DCU.

Eoghan O’Leary

You will also encounter the Key to Success information boxes. These boxes contain expert analysis, explanations, and hints and tips from the authors that will help you maximise your grade in the exam.

Ciarán Duffy

Since 2018, Ciaran has been Deputy Principal of St Kevin's Community College, Clondalkin. He has lectured Junior Cycle mathematics methodologies to trainee teachers in DCU, has corrected all levels of mathematics papers with the SEC since 2002 and has written numerous mock papers and marking schemes. Since 2012, Ciarán has served on the committee of the Dublin Branch of the Irish Mathematics Teachers' Association and has been national chairperson of the IMTA since 2019.

Eoghan is an Assistant Principal and teacher of Mathematics and Economics. Eoghan was shortlisted for the Award for Irish Teachers of Mathematics in both 2017/18 and 2018/19. He is a former Chairperson of the Cork Maths Teachers association. Eoghan is also the Founder of and Director of Education at The Tuition Centre, one of the most popular and best-regarded sources of online tuition for secondary school students in Ireland.

New Junior Cycle Grading System

The following are the general instructions found on the exam paper: “There are [x] questions on this examination paper. Answer all questions. Questions do not necessarily carry equal marks. To help you manage your time during this examination, a maximum time for each question is suggested. If you remain within these times you should have about 10 minutes left to review your work.”

Time allocation: We recommend that you follow the timings as given on the exam paper. It’s a good idea to jot down, on the exam paper, your planned end time for each question as you go along. We strongly advise you to practise past papers and sample papers of this type to get used to this kind of time management. Try doing a full paper under exam conditions and with a time limit of 2 hours, and see how that goes for you. It is essential that you answer all questions, so make sure you do not run out of time.

Answer all questions.

Always show your work. Some questions will not require this. These questions usually say ‘write down…’, otherwise supporting work is always needed.

One Paper (2 hours) – 270 marks

Put everything on the exam booklet. While you are entitled to as much extra paper as you like, we recommend that you use the exam booklet itself to do all of your work. There is no such thing as ‘rough work’ – all work is part of your solution.

is the front cover of your exam. You fill in your exam number and the day and month of your birth ONLY. Never write your name or school on ANYTHING.

Understanding How the Marking Process Works

Your examiner must read everything you do. If you try a question two different ways, your examiner will have to correct both and give you the higher mark of the two attempts. So, don’t scribble things out, don’t use Tippex, don’t use pencil (so you can’t rub things out).

Consider the example below.

(a) Jane buys a laptop online for $699, plus a shipping cost of $30 The exchange rate is $1 = €0.90 Work out in euro the total cost to Jane of buying the laptop online.

7 10 4 10 10 10

699 + 30 = 729 - 0.9 = €810

699 x 0.9 = €629.10

699 + 30 = 729 x 0.9 = €656.10

If all three attempts are present in your answer box, then you will get the marks for the best solution. This does not apply, however, when asked a question with only two choices as shown below.

The ratio of students to teachers in Liam's school was 15:1. The school hired one extra teacher, while the number of students stayed the same. The new ratio of students to teachers was a:1 where a ∈ Q.

(c) Put a tick ( ) in the correct box to show shich statement is true. Tick one box only. Give an example to support your answer.

a > 15 a = 15 a < 15

Example:

If you tick more than one of the boxes then you will get 0 marks. If you make a mistake, just make it clear to your examiner which one is your answer. Maybe put a circle around it and say ‘this is my answer’ beside it and/or cross out the wrong answer. This is the only time you should ever scribble out work. Otherwise, put a light line through your work and do it again on the side. There should be plenty of room.

Mark Allocation: Part a) is usually easier than part b), which in turn is usually easier than part c), etc. However, there is no fixed allocation of marks across parts a), b), c), etc. In fact,

Other Top Tips

Diagrams: Drawing a diagram can be really helpful to visualise something that is difficult. You don’t need to be told to draw one, but doing so can make your life easier. If there is a diagram given in the question, it can be useful to draw in any known measurements or angles, particularly 90° angles. Be careful not to assume something is a right angle.

Right Angles: Students often mistake angles as being right angled or 90° when they are not. You must be told it is a right angle or given some additional information to deduce that it is. This kind of information includes:

> Perpendicular lines

> Angle of elevation

> Angle in a triangle in a semi-circle (must go through the centre point to be a semi-circle).

> Real-life scenarios in trigonometry such as walls, trees, etc., which we are allowed to assume are perpendicular to the ground.

> If a triangle’s measurements satisfy a2+b2=c2, then the triangle must be right-angled. It is a major error to assume a triangle is right-

many questions have more than three parts. Don’t assume that the later parts of the question will carry the highest marks. Marks may possibly be allocated more heavily towards the earlier parts of some questions. However, this is not decided until the final marking schemes are approved by the chief examiner, long after the exams are over. You simply can’t make any assumptions about where the marks will be allocated! So the recommended approach is to take care to pick up all marks on the easy stuff, and if something seems difficult, at least have a go at it. “Attempt” marks can contribute a lot to your grade.

angled, so be very careful.

Blank Boxes: No matter what the question is, write something down. If you have no idea how to do it, look at the absolute basic principles of the question and see what can be done.

Examples include:

> If you see brackets, multiply at least two things together.

> Label a point as (x,y) or (x1,y1) and (x2,y2) if given two points.

> Draw a right-angled symbol on a diagram.

> Write down a formula and change one letter to a given value.

> Write down an answer from a previous part which might be useful in the new box.

> Is a percentage given? Multiply AND divide by any other number in the question.

> Is there a decimal or fraction or percentage mentioned? Change it to a different form using your calculator

eg 3 4 = 75% = 0.75

> Can you identify a radius or a diagram or half a diameter to get a radius?

> In probability if a fraction is given, take it away from 1 on your calculator.

> Can you do something with the opposite of the situation given?

> Indicate two equal sides and/or angles in a triangle if the word Isosceles is mentioned.

> If you see a horrible-looking expression and a random number elsewhere, just sub the number in. Somewhere, anywhere!

Read the questions carefully: Take in as much of the information given and crosscheck with any diagrams to see if it backs up your understanding of the problem posed. If not, allow the diagram to correct your misunderstandings and proceed from there. Every word on the paper is there for a reason – use them all wisely. At the same time, if the examiner needed to tell you something that you think is missing, then maybe your thought was incorrect. Do not make assumptions (*except in real-life right-angled situations mentioned earlier).

Does your answer make sense? Are you working on a probability question and get an answer bigger than 1? This is not possible, so you must have done something wrong. Or, did you work out a rate of tax to be 75% or an hourly rate of pay to be €1.36 or €235,057? Do the three angles in your triangle add to more than 180° or does the height of a person work out to be 5 metres tall? Check that everything makes sense and have an idea what the answer should/could be or not be. Knowledge and skills that you have acquired outside the classroom are valid and to be used where possible.

It is worth looking at marking schemes from previous years in order to gain an understanding of what marks tend to be awarded for. The process won’t change that much in the future.

That’s what the calculator said: Calculators are operated by humans! If your monthly tax bill is €20 it is likely you’ve forgotten a zero or two somewhere along the way. Be familiar with your own calculator and how it works. Practise all of your classwork and homework with the same calculator where possible. Do you know how to use your calculator to do the following?

> Find prime factors

> Use table mode for drawing graphs

> Change fractions to percentages to decimals

> To work with time (hours and minutes)

> Use Sin, Cos and Tan for ratios and their inverses for angles

> Square root and squaring

> Cubed root and cubing

> Mixed fractions

> Change angle mode from Degrees to Radians (this is only important if you share your calculator with a student in 5th or 6th year). Radians mode is not a concept on the JC course but it is on the LC HL course and could mess up all of your trigonometry answers if you don’t know how to change it.

> Know how to get π on your calculator.

Show work: Correct answers without work usually earn high partial credit (approx. 70% of the marks). Any calculation or step shown before the answer will count as showing work. If a question asks you to label a point or use your graph to get an answer, then you must do so for full marks.

Formula & Tables Book: Be familiar with all of the relevant pages in the

tables book. Reading the contents page can be very helpful. It is not a good idea to use a newly discovered formula on the day of an exam that you have never used in class before. There is probably a reason you have never seen it before!

Order of Questions: Do questions in any order you like but as you have to do them all there’s not a huge advantage to moving back and forth. Some people like to start with questions they like to settle themselves into an exam; that choice is up to you. There is a danger that you skip some pages and miss questions by moving back and forth. The choice is yours!

I’ve never seen this before: You will come across questions that look unusual and nothing like questions you’ve practised in class. This is the same for everyone doing the exam, so take it on the chin and do your best. Apply all of your mathematical knowledge and skills to this unforeseen question and do your best. Most mathematicians don’t know if the method they have chosen to start a question is the correct one. This only becomes apparent after you’ve started. So, start the question and see what happens. If you get inspired half way through to do it differently, then do so. If you run into a brick wall, change your starting position or your tactics and see if something new emerges.

Units: Ensure you are working in the same units when solving some problems involving time, distance, speed, interest, money, etc. If two different units are given in a question, it is a good idea to check what form the answer should be in. Then change the other unit to this. E.g. if we are told the length of a rectangle is 25 cm and the width is 43 mm and we want to know the area in cm2, then we change the mm into cm. Of course, it can be done the other way, but this involves two conversions, is more time consuming and way more prone to error. It is also important to include the correct units as part of your final answer. You can lose 1

mark every time you leave them out. This doesn’t sound like a lot, but they add up and why do it to yourself after doing all the hard work so well?

But I got the right answer: Maths isn’t about answers weirdly enough. It is all about method and process. It is possible to get the correct answer doing something completely illegal, mathematically speaking!

Example 1: Simplify 26 65

The correct answer is 2 5 after you divide top and bottom by 13 However, you get the same answer if you cross out the 6 on top and bottom which of course is extremely hurtful to all of the laws of mathematics and would result in 0 marks in an exam.

Example 2: Evaluate (25)(92)

If the student says (25)(92) = 2592 instead of saying (25)(92)=32×81=2592, they would be correct but they’ve totally ignored the powers. Coincidentally, it is the correct answer.

Example 3: Simplify x2-y2 x-y

If a student does this by "cancelling" the x's and the y's on top and bottom, to get: x-y -

and then concludes that "two negatives make a positive", so the final answer has to be x+y, this gives a correct answer but uses really offensive mathematics!

Stop focusing on the answer. Your method and work are the most important thing in an exam. The best way to approach this is to practise showing work as often as you can in all maths problems you work on.

But I don't know how to study maths! The best way to study maths is to do maths. Find questions, even ones you've done before and practise, practise, practise!

MATHS EXAM LANGUAGE

Give an account of the similarities and (or) differences between two (or more) items or situations, referring to both (all) of them throughout

Using only the prescribed tools of compass, set square, ruler, pencil and/or protractor to accurately draw properties of shapes and geometric results

Change from one form to another

State or calculate a rough value for a particular quantity

Give a reasoned account, showing how causes lead to outcomes

Judge the relative quality or validity of something, which may include analysing, comparing and contrasting, criticising, defending or judging. When working with numbers it can also simply mean - find the value of.

Use the answer from a previous part to solve this part. If you don’t, you can still get high marks, just not full marks.

We suggest you use the answer from the last part to solve this part, but if you can find a different way then that’s ok for full marks.

Use knowledge and understanding to explain the meaning of something in context

Give valid reasons or evidence to support an answer or conclusion

Give a deductive argument to demonstrate that a particular statement is true, including reasons for each step in the argument.

Give the number in the required form (for example, a multiple of 10, or a number with two significant figures) that is closest in absolute terms to a particular number.

Any correct method can be used to show that two things are the same (except measuring something with a ruler unless specified).

Draw a rough diagram or graph without using geometrical tools.

Provide a concise statement with little or no supporting argument.

To get full marks you must show the examiner that you have used the graph given or that you have drawn. This is best done by drawing feint lines on the graph and marking clearly any points of intersection of your line and the graph.

Little or no supporting work is required.

Demonstrate that a statement is true

WORKING WITH PATTERNS

Topic 4 SAMPLE

Introduction

Linear Sequence

A pattern that is given as an ordered list is called a sequence and a pattern of numbers is called a number sequence. A number sequence is called LINEAR when it forms a straight line when graphed on a co-ordinate plane.

4, 9, 14, 19, ... is an example of a number sequence. Each number in the sequence is called a Term. We use the notation T1 to represent the 1st term, T2= 2nd term etc.

NotationWords

T1 = 4 the first term is 4

T2 = 9 the second term is 9 T3 =14 the third term is 14 T4 =19 the fourth term is 19

The difference between each term is called the first difference.

T1 T2 T3 T4 Value of Term 49 1419 1st Difference +5+5 +5

Real-world application: Alisha measures a plant in her house as part of her CBA. It is 4 cm tall. She predicts it will grow by 5 mm each week and draws a graph of what the height of the plant would be after 4 weeks.

It is easy to work out what the 5th term is, even the 10th term, by continuing the pattern. But what if we wanted the 100th term or the 3000th term? It would take too long to write out the full sequence. This is why we need a rule to find any term. This is called the general term of a sequence

General Term of a Linear Sequence

Method 1 Method 2

Using the formula T n = a + (n − 1) d. This is on page 22 of the Formulae and Tables book. a = the 1st term n = any term you want in the sequence d = 1st difference/common difference

The first difference is 5. The general term will always start with 5n. We use inspection to figure out what else has to happen to get the sequence we have.

Find the general term (or T n) of the sequence 4, 9, 14, 19…

a = 4 and d = 5 T n = 4 + (n −1) (5) T n = 4 + 5n 5 T n = 5n − 1

The 5 is being multiplied by both the 'n' and the '−1'

5n means 5 multiplied by the term number. The first term (when n = 1) needs to give us 4 according to our sequence 5 (1) = 5. What do we need to do to get to 4? ⸫ Minus 1

So, our general term is 5n − 1. Try this with a few of the other terms to see if you are correct.

Quadratic Sequence

A sequence where the second differences are the same is called a quadratic sequence. For example: 4, 6, 10, 16, 24, 34, 46,…. is a quadratic pattern.

Examine the differences between the terms. 4610162434

+2 + 4 + 6 +8 +10 +2 +2 +2 +2 The 2nd difference is the same. This means it is a quadratic pattern.

Finding the general term of a quadratic pattern

This is not a linear pattern as the 1st difference is not the same. Now examine the 2nd differences.

We can use a formula to work this out. Quadratic expressions are always of the form

T n = an2 + bn + c 2a = 2nd difference 2a = 2; a = 1 T n = (1) n2 + bn + c T n = n2 + bn + c Now use the first two terms from the given sequence to create two equations which we solve simultaneously.

T1 = 4 and T2 = 6 (1)2 + b (1) + c = 4 1 + b + c = 4 b + c = 3 1

and (2)2 + b (2) + c = 6 4 +2b + c = 6 2b + c = 2 2

Get b or c on its own from equation 1 and sub into equation 2 c = 3 − b c = 3 − (b) c = 4

2b + 3 − b = 2 2b − b = 2 − 3 b = −1

General term of this sequence is n2 - n + 4

You may have been told before that the second difference is always 2a, but why is this? Examine the first four terms of the general pattern (replace n with 1, 2, 3 and 4 and see what happens)

T1 = a(1)2 + b(1) + c ⸫ a + b + c T2 = a(2)2 + b(2) + c ⸫ 4a +2b + c T3 = a(3)2 + b(3) + c ⸫ 9a+3b+c T4 = a(4)2 + b(4) + c ⸫ 16a+4b+c So the pattern can also be written as: a + b + c, 4a + 2b + c, 9a + 3b+c, 16a + 4b + c …

Find the 1st difference: T2 − T1 (4a + 2b + c) − (a + b + c) = 3a + b T3 − T2 (9a + 3b + c) − (4a + 2b + c) = 5a + b T4 − T3 (16a + 4b + c) − (9a + 3b + c) = 7a + b

1st differences: (3a + b) , (5a + b) , (7a + b), 2nd difference: (5a + b) − (3a + b) = 2a and (7a + b) − (5a + b ) = 2a

As expected, the second difference is constant (the same). So we can conclude that the second difference in a quadratic sequence is always 2a

Exponential Sequence

A sequence is exponential if each term is multiplied by the same number. This is called a common ratio. For JC HL maths, we just focus on doubling (multiply by 2) and trebling (multiply by 3).

Examples of exponential sequences include 5,15,45,135,405,… 2 3 , 4 3 , 8 3 , 16 3 , 32 3 ,… 3, 6, 12, 24, 48,…

Worked Example 1

Investigate whether the pattern in the table below is linear, quadratic or exponential. Explain your conclusion.

Term 1 Term 2 Term 3 Term 4 Term 5

2a − b + 2c 8a − 2b + 2c 18a − 3b + 2c 32a − 4b + 2c 50a − 5b + 2c

Solution

If linear, then our 1st differences will be constant.

Term 2 − Term 1 = Term 3 – Term 2 Or T2 − T1 = T3 T2

(8)(−18) = −144

If quadratic, then our 2nd differences will be constant. If exponential, then each term is being multiplied by the same value.

Term 2 ÷ Term 1 = Term 3 ÷ Term 2 Or T2 T1 = T3 T2

Check for linear by finding differences between terms. T2 − T1 (8a − 2b + 2c) − (2a − b + 2c) 8a − 2b + 2c − 2a + b 2c =6a − b

2a − b + 2c 8a − 2b + 2c 18a − 3b + 2c 32a − 4b + 2c 50a − 5b + 2c

T3 − T2 (18a − 3b + 2c) − (8a − 2b + 2c) 18a − 3b + 2c − 8a + 2b 2c = 10a − b

T4 − T3 32a − 4b + 2c − (18a − 3b + 2c) 32a − 4b + 2c − 18a + 3b 2c = 14a − b

1st differences are not the same ⸫ not linear. Now we investigate the 2nd differences to see if it’s quadratic.

10a − b − (6a − b) 10a − b − 6a + b 4a

14a − b − (10a − b) 14a − b − 10a + b 4a

As the second difference is the same, it must be a quadratic sequence

Worked Example 2

Gertie writes down the following sequence, which repeats every three terms:

3, 6, 4, 3, 6, 4, 3, …

The 1st term is 3

(a) (i) Write down the value of the 12th term. (ii) Work out the value of the 100th term in this sequence.

(b) Describe how to find the value of the nth term in the sequence, where n ∈ ℕ, without listing all the terms from the 1st to the nth

Gertie made her sequence 3, 6, 4, 3, 6, 4, 3, … by picking 3 as the 1st term, and then using this rule:

If a term is odd, multiply it by 2 to get the next term.

If a term is even, add 2 to it and half your answer to get the next term.

For example, 3 is odd, so the next term is 2 × 3, which is 6 6 is even, so the next term is 1 2 × (6 + 2), which is 4

(c) A different sequence follows the same rule, but has 8 as the 1st term. Work out the next four terms of this sequence.

(d) Ahmed takes 2 as his 1st term and makes a sequence using the same rule State what is unusual about Ahmed’s sequence, by working out some of the terms of his sequence.

(e) In another sequence using the same rule, the 2nd term is 86 Work out the two different values that the 1st term could have in this sequence

(f) A different sequence following the same rule starts with the number k, which is odd Work out the next three terms of this sequence, writing each term in simplest form in terms of k

Solution

(a) (i) Every 3rd term in the sequence is the number 4. All terms that are multiples of 3 will be the number 4. 12 is a multiple of 3 ⸫ 4 is the 12th term. You could also just continue to write out the sequence until you get the 12th term. This is not very practical, as the next part shows.

(ii) As 100 is not a multiple of 3, we find the remainder when we divide 100 by 3. 100 divided by 3 is 33 with a remainder of 1. A complete pattern has 3 terms. Another way of looking at it is saying that the 99th term (i.e. 3 × 33) will be the end of a complete pattern (the number 4 in this question), so the next term (100th term) will be the next number in the pattern, i.e. 3

(b) This was asked in a JC HL exam. This is what the marking scheme was looking for:

Divide n by 3

If remainder = 0 then it is the third term.

If remainder = 1 then it is the first term.

If remainder = 2 then it is the second term.

It was worth 5 marks and students got 2 out of 5 if they mentioned ‘divide by 3’. It was very poorly answered.

Note the use of the word ‘describe’ in the question. Lots of students tried to create a formula for T n, whereas the examiners were looking for a description of the process. Look out for this in future questions.

(c) Term 8 5 10 6 4

Work 1 2 (8 + 2) = 5 5 × 2 = 10 1 2 (10 + 2) = 6 1 2 (6 + 2 ) = 4 etc.

(d) If 2 is the first term, Ahmed must add 2 and then half it to get the next term. 1 2 (2 + 2) = 2 but he gets 2 again, so every term of the sequence will be constant at 2

Scan for Video

(e) Answer: 43 , 86 , or 170 , 86 , ... There are two different values depending on whether the 1st term was even or odd.

If odd: If even: T1 × 2 = 86 T1 = 86 2 T1 = 43

1 2 × (T1 + 2) = 86 T1 + 2 = 2(86) T1 = 172 2 T1 = 170

etc.

(a) The first three terms of a sequence are: Term 1: Term 2: Term 3: 6x − 3 x2 − 2x 4x2 + 3x Express the first differences in their simplest form, in terms of x. (b) Show that, if the terms form a linear sequence, then 2x2 + 13x − 3 = 0.

Solution

Remember, put the second term first.

(b) In a linear sequence, the difference between terms is the same.

3 T2 = T2 T1

2 − 8x + 3 = 3x2 + 5x

T2 = x2 − 8x + 3

T1 = 3x2 + 5x

QED = Quad erat demonstrandum (Latin) It means you have shown (demonstrated) what you were asked to.

2 − 3x2 − 8x − 5x + 3 = 0

A minus outside a bracket causes lots of problems for students. Revise it – understand it – practise it. If it’s helpful, put in the invisible 1 after the minus so it is clear you are multiplying all terms in the bracket by − 1. So it becomes x2 − 2x 1(6x − 3) Multiply both sides by − 1

2x2 − 13x + 3 = 0 2x2 + 13x − 3 = 0

Worked Example 5

The first three stages in a pattern of green and white tiles are shown in the diagram below.

Stage 1 Stage 2 Stage 3 Stage 4

(a) Draw the next stage of tiles (stage 4).

(b) Based on the pattern shown, complete the table below.

Stage (n)

Number of White Tiles Total 1 4 6 2 3 4 5

Number of Green Tiles

Assuming the pattern continues, the total number of tiles in stage n  is given by the formula T n = n2 + bn + c, where b and c ∈ ℕ

(c) Find the value of b and the value of c

(d) Find the number of the stage which has 443 tiles in total.

Solution

(a) Stage 4

(b)

Stage (n) Number of Grey Tiles

Number of White Tiles Total

1 4 2 6 2 9 2 11 3 16 2 18 4 25 2 27 5 36 2 38

(c) 6, 11, 18, 27 ... 5, 7, 9 ... 2, 2 ...

General Formula of a Quadratic Sequence

T n = an2 + bn + c Where 2a = 2nd difference

Let the 2nd difference equal 2a = 2 a = 1

We know that the 1st term of the sequence (when n = 1) is 6.

Sequence 1st difference 2nd difference a = 1 b = 2 c = 3

Update the formula for the general term by subbing in a = 1. T n = n2 + bn + c

T1 = 6 T n = n2 + bn + c (1)2 + b(1) + c = 6 1 + b + c = 6 b + c = 5

We know that the 2nd term of the sequence (when n = 2) is 11.

T2 = 11 T n = n2 + bn + c (2)2 + b(2) + c = 11 4 + 2b + c = 11 2b + c = 7

Solve the simultaneous equations. 2b + c = 7 b + c = 5 b = 2 b + c = 5 2 + c = 5 c = 3

Finally, sub in the values for a, b, and c into the General Formula. T n = n2 + bn + c T n = n2 + 2n + 3

Worked Example 6

(a) Only one linear pattern begins with 1, 7.

Fill in the three boxes below so that the numbers form this linear pattern.

Linear pattern: 1, 7, , ,

(b) Many different quadratic patterns begin with 1, 7

Fill in the three boxes below so that the numbers form a quadratic pattern. Quadratic pattern: 1, 7, , ,

Solution

(a) Linear pattern: 1, 7, 13 , 19 , 25

In a linear pattern, the difference between each term is constant (the same). 1st difference = 7 1 = 6

Keep adding 6 to the previous term to get the next term in the sequence.

(b) Quadratic pattern: 1, 7, 15 , 25 , 37

1st difference 6 8 10 12 2nd difference 2 2 2

In a quadratic pattern, the 2nd difference is constant (the same). In this pattern, let the 2nd difference equal to 2 (you can pick anything for this question though) so that the difference between the terms increases by 2 each time. The third term is 8 larger, the fourth is 10 larger, etc.

Scan for Video

Worked Example 7

(a) In a particular linear sequence, the second term is 40 and the sixth term is 116 Fill in the boxes below to show the rest of the first six terms of this sequence.

? 40 ? ? ? 116 ,,,,,

(b) Orla is asked to write down a quadratic sequence. She writes down the following: 5, 6, 9, 14, 22, 30, 41 Exactly one of the terms in Orla’s sequence is incorrect.

Find the correct quadratic sequence by only changing one of the terms in Orla’s sequence.

Solution

(a) As the sequence is linear, we know the difference between each term is the same. ? , 40 , ? , ? , ? , 116

So the gap from 40 to 116 is 4 times the common difference. 116 40 = 76, which is 4 jumps. Divide 76 by 4 to get the common difference. 76 ÷ 4 = 19 So, the number to the left of 40 is 40 19 = 21 and we can now fill in the rest of the sequence. 21 , 40 , 59 , 78 , 97 , 116 19 19 19 19 19

(b) In a quadratic sequence, the second difference is constant (the same). If there can only be ONE incorrect entry, then it must come from the 22

Sequence 5 6 9 14 22 30 41

1st difference 1 3 5 8 8 11

2nd difference 2 2 3 0 3

For there to be a 2nd difference, there has to be a 1st difference initially. So, the 22 must be wrong. Sequence 5 6 9 14 21 30 41

1st difference 1 3 5 7 9 11 2nd difference 2 2 2 2 2

John makes a sequence where each stage is made up of a certain number of X's arranged in a pattern. The first three stages of John's sequence are shown below. The sequence starts at stage 0

Stage 0 Stage 1 Stage 2

(a) Draw the next stage of John's sequence.

(b) Using a table, a graph, or otherwise, write a formula to express N in terms of S, where N is the number of X's in stage S of John's sequence.

(c) There are exactly 130 X's in stage k of John's sequence. Find the value of k.

(d) Yoko is also making a sequence, with each stage made up of a certain number of X's arranged in a pattern. In Yoko's sequence, the relationship between N and S is given by the formula:

N = 1 + 2S where N is the number of Xs in stage S of the sequence (starting at stage 0).

(i) Draw one possible example of the first three stages of Yoko's sequence in the table below.

Stage 0 Stage 1 Stage 2

(ii) p represents the number of X's in stage y of Yoko's sequence. Write down the number of X's in stage y + 3 of Yoko's sequence. Give your answer in terms of p

(d) (i)

X X X X X X X X X

Stage 0 Stage 1 Stage 2

There are many different correct answers to this question. The main thing you need to have is one X in Stage 0 and an extra two X's for each of the following stages.

(ii) N = 1 + 2S and N is the number of X's in Stage S.

It is useful to write out in words and algebra what’s going on here. The number of X's in Stage y is = p.

p = 1 + 2y N = 1 + 2(y + 3) N = 1 + 2y + 6 and (1 + 2y) is p from above N = p + 6 So, the number of X's in stage y + 3 is (p + 6).

Over to You

Q1 Pete is saving to buy an Xbox. Pete has saved €20 to begin with. He saves a further €12 each week. Find the total amount of money Pete will have saved after 5 weeks.

Q2 84 112 144

These triangles can be put in a sequence of increasing size. The lengths of the bases of the triangles in this sequence follow a quadratic pattern. Three consecutive triangles in this sequence are shown below. Use this information to find the length of the base of the next triangle in the sequence.

The length of the hypotenuse, h, of triangle x in this sequence is given by the function below, where b and c are integers. h(x) = 2x2 + bx + c Also, h(1) = 5 and h(2) = 13. Use this information to write two equations in b and c, and solve to find b and c.

Q3

A boxer runs up stairs as part of her training.

She can go up 1 step or 2 steps with each stride, as shown.

The boxer wants to count how many different ways she can reach the nth step.

She calls this T n , the nth Taylor number.

For example, she has three different ways to reach the 3rd step, as shown in the tables below.

So T 3 = 3.

(a) Find the value of T1 and T2

Up 1 step

3rd step: way 1 up 1 step then 1 step then 1 step 1 + 1 +1

Up 2 steps

3rd step: way 2 up 1 step then 2 steps 1 + 2

3rd step: way 3 up 2 steps then 1 step 2 + 1

(b) List all the different ways that she can reach the 4th step; one way is already done for you. Hence, write down the value of T4

(c) Some of the ways to reach the nth step start by going up 1 step; others start by going up 2 steps.

(i) List the different ways that she can reach the 5th step, if she starts by going up 1 step.

(ii) List the different ways that she can reach the 5th step, if she starts by going up 2 steps.

Q4 The first three stages of a pattern are shown below. Each stage of the pattern is made up of small squares. Each small square has an area of one square unit.

Stage 1 Stage 2 Stage 3

(a) Draw the next two stages of the pattern.

(b) The perimeter of Stage 1 of the pattern is 4 units. The perimeter of Stage 2 of the pattern is 12 units. Find a general formula for the perimeter of Stage n of the pattern, where n ∈ ℕ

(c) Find a general formula for the area of Stage n of the pattern, where n ∈ ℕ

(d) What kind of sequence (linear, quadratic, exponential, or none of these) do the areas follow? Justify your answer.

Q5 The table shows the height, in metres, of a ball at various times after being kicked into the air.

Time (seconds) 00.511.522.53 Height (metres) 0.33.45.77.27.97.86.9

Is the pattern of heights in the table linear, quadratic, or exponential? Explain your answer.

Q6 The first three terms of a linear sequence are −5, k, 1.

(i) Find k and hence, or otherwise, show that the common difference is 3.

(ii) Find T10, the 10th term in the sequence.

(iii) Find which term in the sequence has a value of 247

Q7

(a) The first three patterns in a sequence of patterns containing dots and crosses are shown below.

Pattern 1 Pattern 2 Pattern 3

(i) Draw the fourth pattern in the sequence.

(ii) Find a formula, in n, for the number of dots in pattern n of the sequence (Tn ) (iii) Find the total number of dots in the first 20 patterns of the sequence. The table shows the number of crosses for the first two patterns.

Pattern 123456

Number of crosses 14

(iv) Complete the table, and hence find a formula for the number of crosses in pattern n of the sequence.

(v) Find the number of crosses in pattern 12 of the sequence.

(b) The first three patterns in a sequence of patterns of crosses are shown below.

Pattern 1 Pattern 2

The number of crosses in pattern n is T n . The general term describing T n can be written in the form:

T n = n2 2 + bn + c, where b, c, ε R.

(i) Use substitution for n to write T1 and T2 in terms of b, c, and a number.

(ii) Solve to find the values of b and c.

Q8 Part of the seating arrangement in a theatre is shown in the diagram below. The seats are arranged in rows. Row 1 is nearest the stage and has 28 seats. Each subsequent row behind that contains one more seat. i.e., row 2 has 29 seats, row 3 has 30 seats, and so on.

Stage

(a) Find the number of seats in row 10.

Q9 The first three patterns in a sequence of patterns of tiles are shown in the diagram below.

Pattern 1 Pattern 2 Pattern 3

(a) Draw the next pattern of tiles.

(b) Based on the patterns shown, complete the table below

Pattern number (n) Number of tiles 1 5 2 3 4 5

(c) Assuming the pattern continues, the number of tiles in the nth pattern of the sequence is given by the formula T n = pn + q, where p and q ∈ ℕ.

(i) Find the value of p and the value of q (ii) How many tiles are in the 20th pattern? (iii) Find which pattern has exactly 290 tiles.

Q10 The table below shows the first five terms of an arithmetic sequence. Find an expression for T n, the nth term of the sequence. Hence, or otherwise, find the value of T30, the 30th term.

on