∑ DANEMATHICS
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Grade 12 · Patterns & Sequences
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)constant✓B)zeroC)increasingD)equal to the first differenceAnswer: A — A constant second difference is exactly what makes a pattern quadratic.
- B — a zero second difference makes the pattern LINEAR
- C — an increasing second difference means a cubic pattern or higher
- D — the first differences change; the second ones do not
- 1.2For the pattern 204 ; 176 ; 150 ; 126 ; …, the SECOND difference is …(1)A)2✓B)−28C)−2D)−26Answer: A — The first differences are −28, −26 and −24, and each is 2 more than the one before.
- B — that is the FIRST difference
- C — the first differences are INCREASING, so the second difference is positive
- D — that is the second of the first differences
- 1.3The next TWO terms of 204 ; 176 ; 150 ; 126 ; … are …(1)A)104 and 84✓B)102 and 78C)104 and 82D)100 and 74Answer: A — The first differences continue −22 and then −20.
- B — carried on subtracting 24 each time, ignoring the second difference
- C — used −22 twice instead of −22 then −20
- 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 second difference itselfC)the first differenceD)the first termAnswer: A — The second difference always equals 2a.
- B — the second difference is 2a, so a is half of it
- C — the first difference changes from term to term
- D — the first term is a + b + c, not a
- 1.5The general term of 204 ; 176 ; 150 ; 126 ; … is …(1)A)Tn = n² − 31n + 234✓B)Tn = 2n² − 31n + 234C)Tn = n² − 28n + 231D)Tn = n² + 31n + 234Answer: 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 — used the first difference −28 as b directly
- D — b must be negative, since the pattern falls at first
- 1.6A quadratic pattern with a > 0 has …(1)A)a minimum term✓B)a maximum termC)no turning termD)a constant first differenceAnswer: A — The terms lie on an upward parabola, so they fall and then rise.
- B — a maximum term needs a < 0
- C — the parabola always turns somewhere
- D — a constant FIRST difference would make the pattern linear
- 1.7The general term of 3 ; 8 ; 17 ; 30 ; … is …(1)A)Tn = 2n² − n + 2✓B)Tn = 4n² − n + 2C)Tn = 2n² + n + 2D)Tn = 2n² − nAnswer: A — The second difference is 4, so a = 2; then 3a + b = 5 gives b = −1, and c = 2.
- B — used the second difference 4 as a instead of half of it
- C — 3a + b = 5 with a = 2 gives b = −1, not +1
- D — left out c, which would make T1 equal 1, not 3
- 1.8For Tn = n² − 31n + 234, the SMALLEST term occurs at …(1)A)n = 15 and n = 16, which are equal✓B)n = 15,5C)n = 31D)n = 1Answer: A — The vertex is at n = 15,5, and n must be a whole number, so the two neighbouring terms tie at −6.
- B — n counts terms, so it must be a whole number
- C — 31 is the value of −b, before dividing by 2a
- D — the pattern falls at first, so the first term is not the smallest
- 1.9For Tn = n² − 31n + 234, the number of NEGATIVE terms is …(1)A)4✓B)5C)6D)2Answer: A — n² − 31n + 234 = (n − 13)(n − 18), which is negative for 13 < n < 18, so n = 14, 15, 16 and 17.
- B — included n = 13 or n = 18, where the term is exactly 0
- C — included both endpoints, but a zero term is not a negative one
- D — counted only the two smallest terms
- 1.10The FIRST differences of a quadratic pattern themselves form …(1)A)a linear pattern✓B)a quadratic patternC)a geometric patternD)a constant patternAnswer: A — They change by the constant second difference each time, which is what a linear pattern does.
- B — the SECOND differences would then be changing, not constant
- C — a geometric pattern multiplies by a ratio, it does not add
- D — constant first differences would make the ORIGINAL pattern linear
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.