∑ DANEMATHICS
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Grade 12 · Patterns & Sequences
Series, Sigma Notation & Convergence (Grade 12)
MARKING GUIDELINE
Marks
32
Duration
50 minutes
Questions
3
Name:
Class:
Date:
Mark
/ 32
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.
- 1.1The value of the sum of 2k, for k running from 1 to 5, is …(1)A)30✓B)10C)15D)25Answer: A — 2 + 4 + 6 + 8 + 10 = 30.
- B — gave only the LAST term
- C — added k from 1 to 5 without the factor of 2
- D — left one of the five terms out
- 1.2In sigma notation, the number written BELOW the sigma is …(1)A)the first value of the counter✓B)the number of termsC)the last value of the counterD)the common differenceAnswer: A — The lower limit says where the counting starts; the upper limit says where it stops.
- B — the number of terms is upper − lower + 1
- C — that is the number written ABOVE the sigma
- D — the difference lives inside the expression, not on the limits
- 1.3A sum runs with its counter from k = 3 to k = 20. The number of terms is …(1)A)18✓B)17C)20D)23Answer: A — upper − lower + 1 = 20 − 3 + 1 = 18.
- B — forgot the + 1, which counts the starting term itself
- C — used the upper limit on its own
- D — ADDED the two limits
- 1.4A geometric series CONVERGES when …(1)A)−1 < r < 1✓B)r > 1C)r ≥ 1D)r ≤ −1Answer: A — Only then do the terms shrink towards zero.
- B — the terms grow, so the sum has no limit
- C — r = 1 makes every term the same, so the sum grows without limit
- D — the terms grow in size and alternate in sign
- 1.5The sum to infinity of 12 + 6 + 3 + … is …(1)A)24✓B)21C)12D)48Answer: A — S∞ = 12 ÷ (1 − 0,5) = 24.
- B — added the first three terms and stopped
- C — that is the first term
- D — used 1 − r as 0,25 instead of 0,5
- 1.6For which values of x does 1 + x + x² + x³ + … converge?(1)A)−1 < x < 1✓B)x > 1C)x is not 0D)all real xAnswer: A — The common ratio is x, so convergence needs x to lie strictly between −1 and 1.
- B — for x > 1 the terms grow and the sum diverges
- C — x = 0 is perfectly fine; it is LARGE values of x that break convergence
- D — for x of size 1 or more the series diverges
- 1.7The value of the sum of 3k, for k running from 1 to 10, is …(1)A)165✓B)55C)30D)150Answer: A — 1 + 2 + … + 10 = 55, and 3 × 55 = 165.
- B — left out the factor of 3
- C — gave only the last term, 3 × 10
- D — used 1 + 2 + … + 10 = 50 instead of 55
- 1.8The value of the sum of the CONSTANT 4, for k running from 1 to 7, is …(1)A)28✓B)4C)7D)11Answer: A — The constant is added seven times.
- B — that is the constant itself, added once
- C — that is the number of terms
- D — ADDED the constant to the number of terms
- 1.9A SERIES is the … of a sequence.(1)A)sum✓B)listC)ratioD)differenceAnswer: A — A sequence lists the terms; a series adds them up.
- B — the list IS the sequence
- C — the ratio belongs to a geometric SEQUENCE
- D — the difference belongs to an arithmetic SEQUENCE
- 1.10If S∞ = 40 and a = 10, then r = …(1)A)0,75✓B)0,25C)4D)30Answer: A — 10 ÷ (1 − r) = 40 gives 1 − r = 0,25, so r = 0,75.
- B — that is 1 − r, not r
- C — that is 40 ÷ 10, which equals 1 ÷ (1 − r)
- D — SUBTRACTED a from the sum instead of solving for r
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
[9 MARKS]Consider the series 3 + 9 + 15 + … + 273.
- 2.1Determine how many terms there are in the series.(3)a = 3 and d = 6 (1)
3 + (n − 1)6 = 273 (1)
n = 46 (1) - 2.2Determine the sum of the series.(3)S46 = 462(3 + 273) (2) = 6 348 (1)
- 2.3Write the series in sigma notation.(3)Tn = 6n − 3 (2)
Σn=146(6n − 3) (1)
Question 3
[13 MARKS]Answer the questions below on geometric series.
- 3.1Determine the value of n if Σk=3n 8(2)k−1 = 131 040.(5)The full sum from k = 1 is 8(2n − 1) (1)
The first two terms are 8 + 16 = 24 (1)
8(2n − 1) − 24 = 131 040 (1)
2n = 16 384 (1)
n = 14 (1) - 3.2Gold is extracted from an old mine heap. Processing 1 000 tons of gravel yields 30 kg of gold the first time, 24 kg the second, 19,2 kg the third and so on. Determine the MAXIMUM amount of gold that can be recovered.(4)r = 2430 = 0,8, and |r| < 1 so the series converges (2)
S∞ = 301 − 0,8 (1) = 150 kg (1) - 3.3Determine the values of x for which the series Σ3(2x)n−1 converges.(4)r = 2x (1)
|2x| < 1 (1)
−1 < 2x < 1 (1)
−12 < x < 12 (1)
TOTAL: 32 marks
This question paper consists of 3 questions.