GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
MATHEMATICS
Grade 11
SESSION 10
(LEARNER NOTES)
QUADRATIC NUMBER PATTERNS
Learner Note: In this session you will revise quadratic number patterns where there is a
constant second difference. The method used is discussed in Section C of this session.
SECTION A: TYPICAL EXAM QUESTIONS
Question 1
(a)
(15 minutes)
Consider the following number pattern: 8 ;18 ; 30 ; 44 ; ........
Determine the next three terms, the general term and the 20th term.
(b)
(6)
Consider the following number pattern:
2;  3;  6; 11;.................
(1)
(2)
Write down the next two terms.
Determine the general term for the number pattern.
(2)
(4)
(3)
Which term of the number pattern is 83 ?
(5)
Question 2
(10 minutes)
Determine which term of the number pattern 8 ;10 ;16 ; 26 ; ........ equals 170.
(9)
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GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
MATHEMATICS
Grade 11
SESSION 10
(LEARNER NOTES)
SECTION B: SOLUTIONS AND HINTS
1(a)
abc
3a  b
12
10
2a
2
T5
T6
T7
60
78
98
16
14
2
18
2
2
20
 60, 78, 98
 a 1
 b7
 c0
Tn  n2  7n
 T20  540

2
(6)
2a  2
a 1
3a  b  10
 3(1)  b  10
abc 8
1  7  c  8
b  7
c  0
 Tn  1n 2  7 n  0
 Tn  n 2  7 n
Tn  n 2  7n
 T20  (20) 2  7(20)
 T20  540
1(b)(1)
6
3
2
27
18
11


18
27
(2)
2
2
7
5
3
1
2
9
2
Next two terms are 18;  27
1(b)(2)
2a  2
3a  b  1
a  b  c  2
a  1
 3(  1)  b  1
1  2  c  2
b  2
 Tn  n  2n  3
2
 c  3
a  1
b2
c  3
2
 Tn  n  2n  3



(4)
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GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
MATHEMATICS
1(b)(3)
Grade 11
SESSION 10
(LEARNER NOTES)
83  n2  2n  3
 Tn  83
 n  2n  80  0
 standard form
 factors
 two values for n
 indicating n  10
2
 0  (n  10)(n  8)
 n  10
or n  8
But n  8
(4)
 n  10
a2
b  4
c  10
2
 Tn  2n  4n  10
 Tn  170



2
abc
3a  b
6
2
2a
4
2a  4
3a  b  2
a2
 3(2)  b  2
 b  4
10
 standard form
 factors
 two values for n
 indicating n  10
4
(9)
abc 8
 2  (4)  c  8
 c  10
 Tn  2n 2  4n  10
170  2n 2  4n  10
 85  n 2  2n  5
 0  n 2  2n  80
 0  (n  10)(n  8)
 n  10
or
n  8
But n  8
 n  10
 T10  170
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GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
MATHEMATICS
Grade 11
SESSION 10
(LEARNER NOTES)
SECTION C: ADDITIONAL CONTENT NOTES
Suppose that the general term of a particular quadratic number pattern is given by
Tn  an2  bn  c .
The terms of the number pattern would then be:
T1  a(1) 2  b(1)  c  a  b  c
T2  a(2) 2  b(2)  c  4 a  2b  c
T3  a(3)2  b(3)  c  9a  3b  c
T4  a(4) 2  b(4)  c  16a  4b  c
abc
4a  2b  c
3a  b
9a  3b  c
7a  b
5a  b
2a
16a  4b  c
2a
First difference
Second difference
You will notice that the constant second difference is given by the expression 2a .
The first term in the first difference row is given by 3a  b
and the first term is given by a  b  c .
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GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
MATHEMATICS
Grade 11
SESSION 10
(LEARNER NOTES)
SECTION D: HOMEWORK
Question 1
A sequence of isosceles triangles is drawn. The first triangle has a base of 2cm and height of
2cm. The second triangle has a base that is 2cm longer than the base of the first triangle. The
height of the second triangle is 1cm longer than the height of the first triangle. This pattern of
enlargement will continue with each triangle that follows.
(a)
Determine the area of the 100th triangle.
(4)
(b)
Which triangle will have an area of 240cm2 ?
(5)
Question 2
Consider the following pattern:
1
2
5
4
7
11
3
8
12
6
9
13
10
14
15
Determine the middle term of the 51st row.
(8)
Page 5 of 6
GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME
MATHEMATICS
Grade 11
Session 10
(LEARNER NOTES)
SECTION E: SOLUTIONS TO SESSION 9 HOMEWORK
Mixed
exercise
(a)
5 ; 8 ;11;14 ; .........
Tn  5  (n  1)(3)
 Tn  5  ( n  1)(3)
 Tn  3n  2
 T16  3(16)  2  50
 T16  3(16)  2  50
Mixed
exercise
(b)
 Tn  3n  2
(3)
1; 2 ; 4 ; 8 ;16 ; .........

Tn  (1)(2) n1
Tn  (1)(2)n1

T16  215
 T16  215
1.1
(2)
17 ; 20 ; 23 ; .........
 answer
(1)
1.2
Tn  8  (n  1)(3)
 Tn  8  (n  1)(3)
 Tn  8  3n  3
 Tn  3n  5
 Tn  3n  5
1.3
T10  3(10)  5  35
(2)
 T10  3(10)  5  35
(1)
2.1
27 ;18 ;12 ; 8 ; .........
 answer
(1)
2.2
2
Tn  27  
3
n 1
 answer
(2)
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