Sample
When you practise the same type of sum or problem over and over, you often get lazy and do not reflect upon the exercise anymore. You are convinced that you know exactly what type of sum or problem you need to solve. But in a test or exam, all these problems are mixed up and then it might be difficult to know what to do.
When mixed exercises form part of your learning process, you learn to identify and complete a sum or problem correctly. This means that you are truly prepared for tests or exams, because you can recall your work instead of merely recognising it.
Self-evaluation
In each theme, and usually following each sub-theme, there is an activity where you need to reflect critically about the extent to which you have mastered certain concepts and procedures. This activity has the following format:
Use the following scale to determine how comfortable you are with each topic in the table below:
1. Alarm! I don’t feel comfortable, but I just need more time to work through the topic again.
2. Help! I don’t feel comfortable with the topic at all. I need help.
3. OK! I feel moderately comfortable with the topic, but I still struggle sometimes.
4. Sharp! I feel comfortable with the topic.
5. Party time! I feel totally comfortable with the topic and can even answer more complicated questions about it.
Complete the table:
Tip: Complete each self-evaluation as honestly as possible. If there are aspects which you have not mastered, revisit these and make sure that you do master them. Ask the facilitator for help. It is important not to move on to a next theme or sub-theme before you have mastered the topic involved, even if this means that you spend more time on a specific theme than recommended by the CAPS.
Assessment criteria
Visit Impaq’s online platform (OLP) for the assessment plan and comprehensive information about the compilation and mark allocation of tests, assignments and examinations. The number of assignments, mark allocation and relative weighting are subject to change.
Note:
The themes covered in the examination papers are subject to change. Always refer to the portfolio book and assessment plan for updated information about the composition of the examination papers.
The two papers at the end of the year are compiled as follows:
Algebra, equations and inequalities (Themes 3 and 7)
Patterns and series (Theme 1)
Finance, growth and decay (Theme 4)
Functions and graphs (Themes 2 and 3)
Differential calculus (Theme 8)
Probability (Theme 12)
(Theme 11)
Analytical geometry (Theme 9)
Trigonometry (Themes 5 and 6)
Euclidian geometry and measurement (Theme 10)
Tip: Make sure that you know which themes are covered in which paper. The themes covered in the examination papers are subject to change. Always refer to the portfolio book and assessment plan for updated information about the composition of the examination papers.
Note:
• No graphing or programmable calculators are allowed (for example to factorise or find the roots of equations). Calculators should only be used to do standard numeric calculations and to verify calculations done by hand.
• Formula sheets are not provided during tests and final examinations in Grade 12.
Supplementary books
Any other books can be used along with this textbook for extra exercises and explanations, including:
• Maths 4 A��rica, available at www.maths4africa.co.za
• The Si��avula textbook, available online for free at www.siyavula.com
• P��thagoras, available at www.fisichem.co.za.
Calculator
We recommend the CASIO fx-82ES (Plus) or CASIO fx-82ZA. However, any scientific, non-programmable and non-graphing calculator is suitable.
Tip: Ensure that you have a suitable calculator.
Sample
Sample
The third term or the term in the third position, is 2
We write T3 = 2
• Linear patterns
The difference between successive terms is constant. For example: 1; 3; 5; … (the common difference is 2).
The first difference changes by a regular amount, and the second difference is constant. For example: 1; 3; 7; 13; … (the second difference is 2).
The general term is given by T n = a n 2 + bn + c
• Exponential equations
How to solve exponential equations of the form
Linear
Revision
number patterns
In a linear number pattern, the general term is T n = dn + c where d = first difference and c is a constant.
Revision example 1:
Determine the general term of a linear number pattern
Find the general term of the number pattern 8 ; 3; 2; …
Solution
Find the first difference between successive terms:
T2 T1 = 3 ( 8) = 5
T3 T2 = 2 ( 3) = 5
The first difference is constant, so the number pattern is linear.
Substitute d = 5 in T n = dn+ c:
T n = 5n + c
To find c, use one of the given terms, e.g. for n = 2 : T2 = 3
T2 = 5(2) + c
3 = 10 + c
c = 13
Substitute in the original equation:
T n = 5n − 13
Revision example 2:
Use the general term of a linear number pattern
The general term of a number pattern is T n = 1 2 n + 3.
a) Find the first three terms of the number pattern.
b) Which term of the number pattern is 40?
Solution
a) T n = 1 2 n + 3
Substitute n = 1, n = 2 and n = 3 in the given formula:
T1 = 1 2(1) + 3 = 3 1 2
T2 = 1 2(2) + 3 = 4
T3 = 1 2(3) + 3 = 4 1 2
The first three terms are: 3 1 2; 4; 4 1 2
b) T n = 1 2 n + 3
Substitute T n = 40 and solve for n:
40 = 1 2 n + 3
1 2 n = 37
n = 2(37)
Theme 1: Patterns, sequence and
n = 74
The 74th term is 40.
Revision example 3:
Sample
Determine the number of terms in a linear number pattern
Given the linear number pattern: 13; 4; − 5; … ; − 113
Determine the number of terms in the pattern.
Solution
First, find the general term.
The first difference is T2 − T1 = 4 − 13
d = 9
T n = dn+ c
T n = 9n+ c
Substitute one of the terms. We will use the first term, T1 = 13.
13 = − 9(1)+ c
c = 13 + 9
c = 22
The general term is:
T n = 9n+ 22
Substitute T n = 113 and solve for n : − 113 = − 9n+ 22
9n = 22 + 113
9n = 135 n = 135 9 = 15
The 15th term is 113.
Quadratic number patterns
In a quadratic number pattern, the general term is: T n = a n 2 + bn + c
Second difference = 2a
First of first differences = 3a + b
First term = a + b + c
Revision example 4:
Determine the general term of a quadratic number pattern
Find the general term of the quadratic number pattern 15 ; 8; 2; 15; …
Solution
Find the first and second differences between the terms of the quadratic number pattern:
First term:
15 = a + b + c
15 = 3 2 + 5 2 + c
∴ c = − 19
Substitute the values of a, b and c in the formula for the general term:
T n = a n 2 + bn + c
The general term of this pattern is T n = 3 2 n 2 + 5 2 n − 19
Revision example 5:
Find the value of a term, given the general term of a quadratic number pattern
Determine the value of the 40th term of the quadratic number pattern with general term
T n = 3n 2 + 3n 12.
Solution
Substitute n = 40 and determine the value of T40:
T40 = 3 (40) 2 + 3(40) − 12
= 4 908
Second difference: 2a = 3
First of first differences:
Revision example 6:
Find the term number, given the general term of a quadratic number pattern
Which term of the quadratic pattern with general term T n = 5 n 2 79n 16 will be equal to 404?
Solution
Substitute T n = 404 and solve for n :
5 n 2 79n 16 = 404
5 n 2 79n 16 404 = 0
5 n 2 79n 420 = 0
n = 20 OR n = − 4,2 (N / A)
∴ n = 20
Thus the 20th term is 404.
Note that n can only be a natural number, so we reject all other solutions.
Revision example 7: Find the next term of a quadratic
Given the following quadratic number pattern, write down the next term: 16; 13; 8; 1; …
Solution
Find the first and second differences between the terms of the pattern and extend the pattern:
Theme 1: Patterns, sequence and
The second difference is 2. This provides a clue to the missing term, as the pattern of first differences requires that 7 + 2 = 9, so 1 + 9 = 8. The next term in the pattern is 8.
Revision example 8: Find an unknown term of a quadratic number pattern
3; 12; k; 48… is a quadratic number pattern. Determine the value of k.
Solution
Make a diagram of differences:
Second differences are equal, so k 21 = 60 2k
Solve for k: 3k = 81 k = 27
Revision exercise
1. Find the general term of the following number patterns: 1.1 33 ; 55; 77; … 1.2 − 30; 50; 130; …
1.3
2. The general term of a number pattern is T n = 1 5 n + 2.
2.1 Find the first three terms of the number pattern.
2.2 Which term of the number pattern is 2?
3. Given the linear number pattern: 17; 36; 55; … ; 473 Determine the number of terms in the pattern.
4. Study the number patterns and calculate the following:
◦ Determine if the sequence is linear or quadratic.
◦ Find the formula for the general term.
◦ Calculate the following three terms in the sequence by using the formula for the general term.
◦ Calculate the 100th term in each case.
4.1 5; 1; 3; …
4.2 1; 4; 9; 16; 25; …
4.3 2; 5; 16; 31; … 5. Given the quadratic number pattern: 2; 3; 5; 8; …
5.1 Write down the next term.
5.2 Determine the general term.
6. Given the following quadratic number pattern: 17; 12; 11; 14; …
6.1 Determine the general term.
6.2 Find the 30th term.
6.3 Which term of the pattern is equal to 182?
7. Give the first three terms of the quadratic number pattern with general term
8. The general term of a quadratic number pattern is T n = 13 n 2 − 5n + 6. Determine the second difference of the pattern.
9. 1; 4; x; 22; … is a quadratic number pattern. Determine the value of x.
10. A quadratic number pattern has general term T n = 3 (n 14) 2 + 8. What is the value of the smallest term of the pattern, and which term has this value?
Solutions
T n = 1 4 n 2 5n + 13 1.1 55 – 33 = 22; 77 – 55 = 22
Linear number pattern with d = 22
130 − 50 = 80
Linear number pattern with d = 80
T n = dn+ c
T n = 80n+ c T2 = 50 = 80(2)+ c c = 50 160 = 110 T n = 80n 110 1.3 − 1 1 4 − (− 7 1 4) = 6 4 3 4 ( 1 1 4) = 6
Linear number pattern with d = 6
T n = dn+ c
T n = 6n+ c
T n = 1 5 n + 2 2.1 T1 = − 1 5(1) + 2 = 1 4 5 T2 = 1 5(2) + 2 = 1 3 5 T3 = 1 5(3) + 2 = 1 2 5 2.2 T n = − 1 5 n + 2 = − 2 1 5 n = 4
n = 20
The 20th term is 2.
3. 36 − 17 = 19; 55 − 36 = 19
Linear number pattern with d = 19
T n = dn+ c
T1 = 17 = 19(1)+ c
c = 2
T n = 19n 2
473 = 19n − 2
19n = 475 n = 475 19 = 25
There are 25 terms in the pattern
4. 4.1 T2 − T1 = − 1 − ( 5) = 4
T3 − T2 = 3 − (− 1) = 4
Theme 1: Patterns, sequence and
Formula for the general term:
T n = dn+ c
T n = 4n+ c
Substitute n = 1 for the 1st term:
Sample
The first differences are equal, so the number pattern is linear and d = 4.
Second differences are constant. Sequence is quadratic. T n = a n 2 + bn + c T1 = 7 1 4 = 6(1)+ c c = − 13 1 4 T n = 6n 13 1 4
T1 = 4(1) + c = 5
Solve for c :
c = 5 4
c = 9
T n = 4n 9
For the 4th, 5th and 6th terms, substitute n = 4; 5; 6:
T4 = 4(4) 9 = 7 T5 = 4(5) 9 = 11 T6 = 4(6) 9 = 15
To find the 100th term, substitute n = 100: T n = 4n 9 T100 = 4(100) 9 = 391 4.2
From 2nd difference:
2a = 2
∴ a = 1
From 1st of the 1st difference:
3a + b = 3
∴ 3(1) + b = 3
∴ b = 0
From given terms: T1 = 1 a + b + c = 1 ∴ 1 + 0 + c = 1
c = 0
Next three terms are:
Second differences are constant. Sequence is quadratic.
T n = a n 2 + bn + c
From 2nd difference:
2a = 4
∴ a = 2
From 1st of the 1st difference:
3a + b = 7 ∴ 3(2) + b = 7
∴ b = 1
From given terms: T1 = 2 a + b + c = 2
2 + 1 + c = − 2
c = 5
Next three terms are: T5; T6; T7 = 50; 73; 100 T100 = 2 (100) 2 + 100 − 5 = 20 000 + 100 5 = 20 095
5.1 Find the first and second differences between the terms of the pattern and extend the pattern:
The second difference is 1. Follow the pattern:
3 + 1 = 4, so 8 + 4 = 12. The next term in the pattern is 12.
5.2 Refer to the diagram of differences: 2 1 1 First difference: Second difference: 2 1 3 3 5 8
Second difference: 2a = 1 ∴ a = 1 2
First of first differences: 1 = 3a + b 1 = 3(1 2) + b 1 3 2 = b b = 1 2
First term: 2 = a + b + c 2 = 1 2 1 2 + c
∴ c = 2
Substitute the values of a, b and c in the formula for the general term:
T n = a n 2 + bn + c
The general term of this pattern is T n = 1 2 n 2 1 2 n + 2
6. 6.1 Find the first and second differences between the terms of the pattern:
Second difference: 2a = 4 ∴ a = 2
First of first differences: 5 = 3a + b
5 = 3a + b
5 = 3(2) + b
b = 11
First term: 17 = a + b + c 17 = 2 11 + c ∴ c = 26
Substitute the values of a, b and c in the formula for the general term:
T n = a n 2 + bn + c
The general term of this pattern is T n = 2 n 2 11n + 26
6.2 Substitute n = 30 and determine the value of T30: T30 = 2 (30) 2 11(30) + 26 = 1 496
6.3 Substitute T n = 182 and solve for n : 2 n 2 11n + 26 = 182 2 n 2 11n + 26 182 = 0 2 n 2 − 11n −
= 12
Thus the 12th term is 182.
7. Substitute n = 1, n = 2 and n = 3:
T1 = 1 4 (1) 2 − 5(1) + 13 = 8 1 4
T2 = 1 4 (2) 2 5(2) + 13 = 4
T3 = 1 4 (3) 2 5(3) + 13 = 1 4
8. a = 13; b = 5; c = 6
Second difference = 2a = 2(13) = 26
9. Make a diagram of differences.
Second differences are equal, so x 7 = 26 2x
Solve for x:
3x = 33 x = 11
10. The minimum value of T n = 3 (n 14) 2 + 8 occurs when n = 14. (Apply quadratic theory: turning point of parabola.)
Minimum value = T14 = 8
1. ARITHMETIC SEQ UENCES (AS)
Sample
A sequence is another name for a number pattern.
In earlier grades you dealt with linear number patterns, in which there is a constant difference between successive terms. Examples include:
1; 3; 5; 7; … The difference is 2, and the general term is T n = 2n − 1 3; 1; 5; 9 ; … The difference is 4, and the general term is T n = 4n + 7
From now we will use the term ‘arithmetic sequences’ for linear number patterns, with a formula for the nth term that looks different but is in fact equivalent to what we used before.
REMEMBER
Formula for the general term of an arithmetic sequence
Let a = first term and d = common difference.
We can therefore write the sequence as:
It is clear that each term in the sequence has the form a + ? d, with a changing coefficient of d. The value of the coefficient of d is one less than the position of the term in the sequence. For example, for the 3rd term, the coefficient of d is 2, which is one less than 3. Therefore, the coefficient of d for the nth term will be one less than n, that is, n 1.
We conclude that the nth term (also known as the general term) of an arithmetic sequence is given by:
T n = a + (n 1)d where a = first term and d = common difference.
The test for an arithmetic sequence is that the difference between any two pairs of successive terms must be the same.
• Revision exercises to refresh prior knowledge.
• Detailed explanations of concepts and techniques.
• Worked examples help learners to better understand new concepts.
• Varied exercises to entrench theory and practise mathematical skills.
• Test papers and memorandums for exam preparation
• Formula sheets and accepted geometrical reasons for quick reference.
• Index of mathematical terms.
• The facilitator’s guide contains step-by-step calculations and answers.
• Use in school or at home.
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