Topic Review — Growth and Decay in Sequences — Solutions

← Growth and Decay in Sequences

This review covers arithmetic sequences, geometric sequences, growth and decay applications, and recurrence relations. Click each answer to reveal the worked solution.

Review Questions

1. Find the 20th term of the arithmetic sequence 7, 12, 17, 22, … ▶ Show Answer
   a = 7, d = 5
   t₂₀ = 7 + 19 × 5 = 7 + 95 = 102
2. Find the sum of the first 15 terms of the arithmetic sequence 3, 7, 11, 15, … ▶ Show Answer
   a = 3, d = 4, n = 15
   S₁₅ = 15/2 × (2 × 3 + 14 × 4) = 7.5 × (6 + 56) = 7.5 × 62 = 465
3. An arithmetic sequence has t₄ = 19 and t₉ = 44. Find a and d. ▶ Show Answer
   a + 3d = 19  (1)
   a + 8d = 44  (2)
   (2)−(1): 5d = 25 → d = 5
   a = 19 − 15 = 4; d = 5
4. Find the 7th term of the geometric sequence 2, 6, 18, 54, … ▶ Show Answer
   a = 2, r = 3
   t₇ = 2 × 3⁶ = 2 × 729 = 1458
5. Find the sum of the first 8 terms of the geometric sequence 1, 2, 4, 8, … ▶ Show Answer
   a = 1, r = 2, n = 8
   S₈ = 1 × (2⁸ − 1)/(2 − 1) = 255 — 255
6. Is 135 a term in the geometric sequence 5, 15, 45, 135, …? Which term? ▶ Show Answer
   a = 5, r = 3
   t_n = 5 × 3ⁿ⁻¹ = 135 → 3ⁿ⁻¹ = 27 = 3³ → n−1 = 3 → n = 4
   135 is the 4th term.
7. A population of 3 200 bacteria grows at 25% per hour. Find the population after 4 hours. ▶ Show Answer
   r = 1.25, a = 3200
   P₄ = 3200 × (1.25)⁴ = 3200 × 2.4414 ≈ 7 813
8. A car is purchased for $32 000 and depreciates at 18% per year. Find its value after 3 years. ▶ Show Answer
   r = 0.82
   V₃ = 32 000 × (0.82)³ = 32 000 × 0.5514 ≈ $17 645
9. Write the first 5 terms of the recurrence relation t_n+1 = t_n + 4, t₁ = 10. What type of sequence is this? ▶ Show Answer
   10, 14, 18, 22, 26
   This is an arithmetic sequence with d = 4.
10. Write the first 5 terms of t_n+1 = 1.5t_n, t₁ = 8. ▶ Show Answer
    8, 12, 18, 27, 40.5
    This is a geometric sequence with r = 1.5.
11. The recurrence relation t_n+1 = 1.06t_n − 3000, t₁ = 40 000 models a loan. Calculate t₂, t₃, and t₄. ▶ Show Answer
    t₂ = 1.06 × 40 000 − 3000 = 42 400 − 3000 = 39 400
    t₃ = 1.06 × 39 400 − 3000 = 41 764 − 3000 = 38 764
    t₄ = 1.06 × 38 764 − 3000 = 41 089.84 − 3000 = 38 089.84
12. An investment of $6 000 earns 4.5% per annum. In how many complete years will it exceed $8 000? ▶ Show Answer
    V_n = 6000 × (1.045)ⁿ
    n = 6: 6000 × (1.045)⁶ ≈ 6000 × 1.3023 = 7814 < 8000
    n = 7: 6000 × (1.045)⁷ ≈ 6000 × 1.3609 = 8165 > 8000 ✓
    After 7 complete years.
13. A geometric sequence has t₂ = 6 and t₅ = 48. Find a, r, and S₈. ▶ Show Answer
    ar = 6 and ar⁴ = 48 → r³ = 8 → r = 2
    a = 6/2 = 3
    S₈ = 3(2⁸−1)/(2−1) = 3 × 255 = 765
14. A sequence is defined by t_n+1 = 0.9t_n + 50, t₁ = 500. Find the first 5 terms and determine the long-term (fixed point) value. ▶ Show Answer
    t₂ = 0.9(500) + 50 = 500 (already at fixed point!)
    Check: L = 0.9L + 50 → 0.1L = 50 → L = 500
    All terms = 500. The initial value equals the fixed point.
15. A theatre has 25 rows. Row 1 has 20 seats; each row has 2 more seats than the previous row. Find: (a) seats in row 25 (b) total seats in the theatre. ▶ Show Answer
    a = 20, d = 2, n = 25
    (a) t₂₅ = 20 + 24 × 2 = 20 + 48 = 68 seats
    (b) S₂₅ = 25/2 × (20 + 68) = 12.5 × 88 = 1 100 seats