10 questions · Form 4 Additional Mathematics Bab 5: Progressions
Find the 15th term of the arithmetic progression: 3, 7, 11, 15, ...
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1. Find the 15th term of the arithmetic progression: 3, 7, 11, 15, ...
Answer: A
a = 3, d = 7 - 3 = 4. T₁₅ = a + 14d = 3 + 14(4) = 3 + 56 = 59.
2. If k + 2, 2k, and 3k + 6 are three consecutive terms of a geometric progression with positive terms, find the value of k.
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).
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.
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 sum to infinity of the geometric progression: 12, 4, 43, 49, ...
Answer: A
a = 12, r = 412 = 13. S_∞ = a1 - r = 12 / (1 - 13) = 12 / (23) = 18.
5. Express the recurring decimal 0.4444... as a fraction in its simplest form using sum to infinity.
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.
6. Find the minimum number of terms of the AP 5, 9, 13, ... required so that its sum exceeds 200.
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.
7. 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.
Answer: A
S_n = (n2)[a + l] => 220 = (n2)[-8 + 52] => 220 = (n2)[44] => 220 = 22n => n = 10.
8. In a geometric progression, T₂ = 12 and T₅ = 324. Find the first term a.
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.
9. The first term of a GP is 5 and the common ratio is 2. Which term of the progression is equal to 320?
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.
10. Calculate the sum of the first 20 terms of the arithmetic progression: 2, 5, 8, 11, ...
Answer: A
a = 2, d = 3, n = 20. S₂₀ = (202)[2(2) + (20 - 1)(3)] = 10[4 + 57] = 10[61] = 610.