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Quiz Chapter 5: Progressions

10 questions · Form 4 Additional Mathematics Bab 5: Progressions

10 questions · plays in fullscreen

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. Find the 15th term of the arithmetic progression: 3, 7, 11, 15, ...

  1. 59
  2. 55
  3. 63
  4. 60
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Answer: A

a = 3, d = 7 - 3 = 4. T₁₅ = a + 14d = 3 + 14(4) = 3 + 56 = 59.

2. Find the common difference of the AP whose n-th term is T_n = 5 - 3n.

  1. -3
  2. 3
  3. 5
  4. -2
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Answer: A

T₁ = 5 - 3(1) = 2. T₂ = 5 - 3(2) = -1. d = T₂ - T₁ = -1 - 2 = -3.

3. A rubber ball is dropped from a height of 10 m. Each time it hits the ground, it bounces back to 45 of its previous height. Find total vertical distance traveled until it stops.

  1. 90 m
  2. 50 m
  3. 80 m
  4. 100 m
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Answer: A

Total distance = Initial drop + 2 × (sum to infinity of upward bounces). Downward = 10. Upward bounces: a = 10(45) = 8, r = 45. S_∞ = 8 / (1 - 45) = 40. Total = 10 + 2(40) = 90 m.

4. Find the minimum number of terms of the AP 5, 9, 13, ... required so that its sum exceeds 200.

  1. 10
  2. 9
  3. 11
  4. 12
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Answer: A

a = 5, d = 4. S_n = (n2)[2(5) + (n - 1)4] > 200 => (n2)[10 + 4n - 4] > 200 => (n2)[4n + 6] > 200 => 2n² + 3n - 200 > 0. For n = 9: 2(81)+27 = 189. For n = 10: 2(100)+30 = 230 > 200. Minimum n = 10.

5. An employee's starting annual salary is RM 36 000 with an annual increment of RM 2 400. Calculate total salary earned after 10 years.

  1. RM 468 000
  2. RM 432 000
  3. RM 504 000
  4. RM 384 000
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Answer: A

AP with a = 36000, d = 2400, n = 10. S₁₀ = (102)[2(36000) + 9(2400)] = 5[72000 + 21600] = 5[93600] = RM 468 000.

6. If k + 2, 2k, and 3k + 6 are three consecutive terms of a geometric progression with positive terms, find the value of k.

  1. 6
  2. 4
  3. 3
  4. 8
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Answer: A

In a GP, T₂ / T₁ = T₃ / T₂ => 2kk + 2 = 3k + 62k => (2k)² = (k + 2)(3k + 6) => 4k² = 3k² + 12k + 12 => k² - 12k - 12 = 0 => (k - 6)(k + 2) = 0... For k = 6: terms are 8, 12, 24 (r = 1.5).

7. Find the common ratio r of the geometric progression: 162, -54, 18, -6, ...

  1. -13
  2. 13
  3. -3
  4. 3
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Answer: A

r = T₂ / T₁ = -54162 = -13.

8. Find the sum to infinity of the geometric progression: 12, 4, 43, 49, ...

  1. 18
  2. 16
  3. 24
  4. 12
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Answer: A

a = 12, r = 412 = 13. S_∞ = a1 - r = 12 / (1 - 13) = 12 / (23) = 18.

9. In a geometric progression, T₂ = 12 and T₅ = 324. Find the first term a.

  1. 4
  2. 3
  3. 2
  4. 6
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Answer: A

T₂ = ar = 12, T₅ = ar⁴ = 324. Divide T₅ by T₂: r³ = 32412 = 27 => r = 3. Since ar = 12 => a(3) = 12 => a = 4.

10. The first term of a GP is 5 and the common ratio is 2. Which term of the progression is equal to 320?

  1. 7th term
  2. 6th term
  3. 8th term
  4. 9th term
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Answer: A

T_n = a r^(n-1) => 5(2)^(n-1) = 320 => 2^(n-1) = 64 => 2^(n-1) = 2⁶ => n - 1 = 6 => n = 7.

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