∑ DANEMATHICS
FREE CAPS MATHS RESOURCES
Grade 11–12 · Number Patterns, Sequences and Series
Quadratic Number Patterns (Grade 11 & 12)
MARKING GUIDELINE
Marks
34
Duration
55 minutes
Questions
3
Name:
Class:
Date:
Mark
/ 34
Instructions and Information
- Answer ALL the questions in this question paper.
- Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
- Show ALL calculations clearly.
- Show all units where applicable.
- Number the answers correctly according to the numbering system used in this question paper.
- A non-programmable calculator may be used, unless stated otherwise.
- Write neatly and legibly.
Question 1
[10 MARKS]Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.
- 1.1In a QUADRATIC number pattern, the second difference is …(1)A)equal to the first differenceB)constant✓C)zeroD)increasingAnswer: B — A constant second difference is exactly what makes a pattern quadratic.
- A — the first differences change; the second ones do not
- C — a zero second difference makes the pattern LINEAR
- D — an increasing second difference means a cubic pattern or higher
- 1.2For the pattern 204 ; 176 ; 150 ; 126 ; …, the SECOND difference is …(1)A)−26B)−2C)2✓D)−28Answer: C — The first differences are −28, −26 and −24, and each is 2 more than the one before.
- A — that is the second of the first differences
- B — the first differences are INCREASING, so the second difference is positive
- D — that is the FIRST difference
- 1.3The next TWO terms of 204 ; 176 ; 150 ; 126 ; … are …(1)A)104 and 82B)104 and 84✓C)102 and 78D)100 and 74Answer: B — The first differences continue −22 and then −20.
- A — used −22 twice instead of −22 then −20
- C — carried on subtracting 24 each time, ignoring the second difference
- D — used −26 twice, going backwards through the differences
- 1.4For Tn = an² + bn + c, the value of a is …(1)A)half the second difference✓B)the first termC)the second difference itselfD)the first differenceAnswer: A — The second difference always equals 2a.
- B — the first term is a + b + c, not a
- C — the second difference is 2a, so a is half of it
- D — the first difference changes from term to term
- 1.5The general term of 204 ; 176 ; 150 ; 126 ; … is …(1)A)Tn = n² − 31n + 234✓B)Tn = 2n² − 31n + 234C)Tn = n² + 31n + 234D)Tn = n² − 28n + 231Answer: A — 2a = 2 gives a = 1; 3a + b = −28 gives b = −31; and a + b + c = 204 gives c = 234.
- B — used a = 2, the second difference itself, instead of half of it
- C — b must be negative, since the pattern falls at first
- D — used the first difference −28 as b directly
- 1.6A quadratic pattern with a > 0 has …(1)A)a constant first differenceB)no turning termC)a minimum term✓D)a maximum termAnswer: C — The terms lie on an upward parabola, so they fall and then rise.
- A — a constant FIRST difference would make the pattern linear
- B — the parabola always turns somewhere
- D — a maximum term needs a < 0
- 1.7The general term of 3 ; 8 ; 17 ; 30 ; … is …(1)A)Tn = 2n² − nB)Tn = 2n² + n + 2C)Tn = 4n² − n + 2D)Tn = 2n² − n + 2✓Answer: D — The second difference is 4, so a = 2; then 3a + b = 5 gives b = −1, and c = 2.
- A — left out c, which would make T1 equal 1, not 3
- B — 3a + b = 5 with a = 2 gives b = −1, not +1
- C — used the second difference 4 as a instead of half of it
- 1.8For Tn = n² − 31n + 234, the SMALLEST term occurs at …(1)A)n = 15,5B)n = 1C)n = 31D)n = 15 and n = 16, which are equal✓Answer: D — The vertex is at n = 15,5, and n must be a whole number, so the two neighbouring terms tie at −6.
- A — n counts terms, so it must be a whole number
- B — the pattern falls at first, so the first term is not the smallest
- C — 31 is the value of −b, before dividing by 2a
- 1.9For Tn = n² − 31n + 234, the number of NEGATIVE terms is …(1)A)5B)4✓C)2D)6Answer: B — n² − 31n + 234 = (n − 13)(n − 18), which is negative for 13 < n < 18, so n = 14, 15, 16 and 17.
- A — included n = 13 or n = 18, where the term is exactly 0
- C — counted only the two smallest terms
- D — included both endpoints, but a zero term is not a negative one
- 1.10The FIRST differences of a quadratic pattern themselves form …(1)A)a linear pattern✓B)a constant patternC)a quadratic patternD)a geometric patternAnswer: A — They change by the constant second difference each time, which is what a linear pattern does.
- B — constant first differences would make the ORIGINAL pattern linear
- C — the SECOND differences would then be changing, not constant
- D — a geometric pattern multiplies by a ratio, it does not add
Answer grid — marking guideline
| 1.1 | A | B | C | D |
| 1.2 | A | B | C | D |
| 1.3 | A | B | C | D |
| 1.4 | A | B | C | D |
| 1.5 | A | B | C | D |
| 1.6 | A | B | C | D |
| 1.7 | A | B | C | D |
| 1.8 | A | B | C | D |
| 1.9 | A | B | C | D |
| 1.10 | A | B | C | D |
Question 2
[12 MARKS]A quadratic number pattern Tn = an2 + bn + c has these properties:
• T1 = −2
• the nth term of the FIRST differences is −6n − 2
• T1 = −2
• the nth term of the FIRST differences is −6n − 2
- 2.1Show that a = −3 and b = 1.(4)The first differences of an2 + bn + c are a(2n + 1) + b (1)
2an + a + b = −6n − 2 (1)
2a = −6, so a = −3 (1)
a + b = −2 gives b = 1 (1) - 2.2Hence determine the general term Tn.(3)T1 = a + b + c = −2 (1)
−3 + 1 + c = −2, so c = 0 (1)
Tn = −3n2 + n (1) - 2.3Determine whether −19 122 is a term of the sequence. Justify your answer with calculations.(5)−3n2 + n = −19 122 (1)
3n2 − n − 19 122 = 0 (1)
Δ = 1 + 4(3)(19 122) = 229 465 (1)
√(229 465) ≈ 479,03, which is not a whole number (1)
n is not a natural number, so −19 122 is NOT a term (1)
Question 3
[12 MARKS]Consider the quadratic pattern 3 ; 8 ; 17 ; 30 ; 47 ; …
- 3.1Show that the pattern is quadratic and determine its general term.(5)First differences 5; 9; 13; 17 (1), second differences 4; 4; 4 — constant (1)
2a = 4 so a = 2 (1); 3a + b = 5 so b = −1 (1); a + b + c = 3 so c = 2
Tn = 2n2 − n + 2 (1) - 3.2Determine which term of the pattern is equal to 122.(4)2n2 − n − 120 = 0 (2)
(n − 8)(2n + 15) = 0 (1)
n = −7,5 is rejected, so it is the 8th term (1) - 3.3Determine the value of the FIRST difference between T20 and T21.(3)The first differences have the rule a(2n + 1) + b = 4n + 1 (2)
At n = 20: 81 (1)
TOTAL: 34 marks
This question paper consists of 3 questions.