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Quiz Chapter 5: Progressions

10 questions · Form 4 Additional Mathematics Bab 5: Progressions

Question 1 of 10Score: 0

The first term of an AP is -8 and the last term is 52. If the sum of all terms is 220, find the number of terms n.

Full Question List & Answer Key

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

1. The first term of an AP is -8 and the last term is 52. If the sum of all terms is 220, find the number of terms n.

  1. 10
  2. 11
  3. 12
  4. 8
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Answer: A

S_n = (n2)[a + l] => 220 = (n2)[-8 + 52] => 220 = (n2)[44] => 220 = 22n => n = 10.

2. Calculate the sum of the first 6 terms of the geometric progression: 3, 6, 12, 24, ...

  1. 189
  2. 192
  3. 381
  4. 93
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Answer: A

a = 3, r = 2. S₆ = 32⁶ - 12 - 1 = 364 - 11 = 3(63) = 189.

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 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.

5. 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.

6. 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.

7. Find the 7th term of the geometric progression: 2, 6, 18, 54, ...

  1. 1458
  2. 486
  3. 4374
  4. 729
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Answer: A

a = 2, r = 3. T₇ = a r⁶ = 2 × 3⁶ = 2 × 729 = 1458.

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. Express the recurring decimal 0.4444... as a fraction in its simplest form using sum to infinity.

  1. 49
  2. 410
  3. 25
  4. 411
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Answer: A

0.4444... = 0.4 + 0.04 + 0.004 + ... which is a GP with a = 0.4, r = 0.1. S_∞ = 0.41 - 0.1 = 0.40.9 = 49.

10. Under what condition does a geometric progression have a sum to infinity (S_∞)?

  1. -1 < r < 1
  2. r > 1
  3. r < -1
  4. r = 1
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Answer: A

A infinite geometric progression converges to a finite sum if and only if |r| < 1, which means -1 < r < 1.

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