Arithmetic Sequences

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Arithmetic Sequences · Grade 12 · Worksheet 1

1. A civil engineer is designing a parabolic arch bridge that spans 80 meters. The arch follows the equation h(x) = -0.02x² + 1.6x, where h is the height in meters and x is the horizontal distance from the left support. To determine the maximum clearance for boats passing under the bridge, the engineer needs to find the maximum height of the arch. What is the maximum height of the bridge arch? Answer: 32 Solution: The height function is h(x) = -0.02x² + 1.6x This is a quadratic function opening downward (a = -0.02 < 0), so the vertex gives the maximum height The x-coordinate of the vertex is given by x = -b/(2a) Substitute a = -0.02 and b = 1.6: x = -1.6/(2 × -0.02) = -1.6/(-0.04) = 40 The maximum height…
   Full step-by-step solution

   Step 1: The height function is h(x) = -0.02x² + 1.6x Step 2: This is a quadratic function opening downward (a = -0.02 < 0), so the vertex gives the maximum height Step 3: The x-coordinate of the vertex is given by x = -b/(2a) Step 4: Substitute a = -0.02 and b = 1.6: x = -1.6/(2 × -0.02) = -1.6/(-0.04) = 40 Step 5: The maximum height occurs at x = 40 meters from the left support Step 6: Substitute x = 40 into the height function: h(40) = -0.02(40)² + 1.6(40) Step 7: Calculate: h(40) = -0.02(1600) + 64 = -32 + 64 = 32 Step 8: The maximum height of the arch is 32 meters The answer is 32.

2. A civil engineer is designing a parabolic arch bridge that spans 80 meters. The arch follows the equation y = -0.02x² + 1.6x, where y is the height in meters and x is the horizontal distance from the left support. To determine the maximum clearance for boats passing under the bridge, the engineer needs to find the maximum height of the arch. What is the maximum height of the bridge arch? Answer: 32 Solution: The arch equation is y = -0.02x² + 1.6x For a quadratic function y = ax² + bx + c, the x-coordinate of the vertex is x = -b/(2a) Here, a = -0.02 and b = 1.6 x = -1.6/(2 × -0.02) = -1.6/(-0.04) = 40 Substitute x = 40 into the equation: y = -0.02(40)² + 1.6(40) y = -0.02(1600) + 64 = -32 + 64 = 32…
   Full step-by-step solution

   Step 1: The arch equation is y = -0.02x² + 1.6x Step 2: For a quadratic function y = ax² + bx + c, the x-coordinate of the vertex is x = -b/(2a) Step 3: Here, a = -0.02 and b = 1.6 Step 4: x = -1.6/(2 × -0.02) = -1.6/(-0.04) = 40 Step 5: Substitute x = 40 into the equation: y = -0.02(40)² + 1.6(40) Step 6: y = -0.02(1600) + 64 = -32 + 64 = 32 Step 7: The maximum height of the arch is 32 meters.

3. Kaia is a construction manager overseeing the laying of a new pipeline. The first section of pipe laid is 12 meters long. Each subsequent section must be 9 meters longer than the previous one to account for the increasing elevation gradient. If the project requires a total of 24 sections to complete the pipeline, what is the length of the 24th section, and what is the total length of the entire pipeline? Answer: The 24th section is 219 meters long, and the total length is 2772 meters. Solution: Identify the first term a1 = 12 meters and the common difference d = 9 meters. Step 2: Use the nth term formula an = a1 + (n-1)d for n = 24: a24 = 12 + (24-1)*9 = 12 + 23*9 = 12 + 207 = 219 meters.
   Full step-by-step solution

   Step 1: Identify the first term a1 = 12 meters and the common difference d = 9 meters. Step 2: Use the nth term formula an = a1 + (n-1)d for n = 24: a24 = 12 + (24-1)*9 = 12 + 23*9 = 12 + 207 = 219 meters. Step 3: Use the sum formula Sn = n*(a1 + an)/2 for n = 24: S24 = 24*(12 + 219)/2 = 24*231/2 = 12*231 = 2772 meters. The answer is that the 24th section is 219 meters long, and the total length is 2772 meters.

4. Emma is training for a charity run and is increasing her weekly running distance by a constant amount each week. In week 1, she runs 7 km. In week 7, she runs 43 km. Her goal is to run at least 100 km in a single week. Assuming the pattern continues, what is the minimum number of weeks she needs to train to reach or exceed 100 km in a week, and what is her total running distance over all those weeks from week 1 up to and including that week? Answer: Week 17, total 867 km Solution: Identify known values. Week 1: a₁ = 7 km. Week 7: a₇ = 43 km.
   Full step-by-step solution

   Step 1: Identify known values. Week 1: a₁ = 7 km. Week 7: a₇ = 43 km. The sequence is arithmetic, so aₙ = a₁ + (n-1)d. Step 2: Use a₇ to find d: 43 = 7 + (7-1)d → 43 = 7 + 6d → 36 = 6d → d = 6. Step 3: Find n such that aₙ ≥ 100: 100 ≤ 7 + (n-1)6 → 100 ≤ 7 + 6n - 6 → 100 ≤ 6n + 1 → 99 ≤ 6n → n ≥ 16.5. Since n must be a whole number, n = 17. Step 4: Verify a₁₇ = 7 + (17-1)6 = 7 + 16×6 = 7 + 96 = 103 km, which exceeds 100 km. Step 5: Find the total distance from week 1 to week 17 using Sₙ = n(a₁ + aₙ)/2: S₁₇ = 17(7 + 103)/2 = 17(110)/2 = 17×55 = 935 km. The answer is week 17, with a total of 935 km.

5. A pharmaceutical company is modeling the concentration of a new drug in a patient's bloodstream over time. The concentration function is given by C(t) = 80te^(-0.2t) milligrams per liter, where t is measured in hours. The company needs to determine the maximum concentration of the drug and the time at which this maximum occurs to establish proper dosing guidelines. Find both the time of maximum concentration and the maximum concentration value. Answer: t = 5 hours, C = 400/e mg/L Solution: C(t) = 80 * t * e^(-0.2 * t) We want to find the time t when C(t) is maximum, and then find the maximum concentration value. Let f(t) = 80t, and g(t) = e^(-0.2t). Derivative of f(t) = 80.
   Full step-by-step solution

   We are given the concentration function: C(t) = 80 * t * e^(-0.2 * t) We want to find the time t when C(t) is maximum, and then find the maximum concentration value. --- **Step 1: Find the derivative of C(t)** We use the product rule: Let f(t) = 80t, and g(t) = e^(-0.2t). Derivative of f(t) = 80. Derivative of g(t) = -0.2 * e^(-0.2t). Product rule: C'(t) = f'(t)g(t) + f(t)g'(t) C'(t) = 80 * e^(-0.2t) + 80t * (-0.2 * e^(-0.2t)) C'(t) = 80 e^(-0.2t) - 16t e^(-0.2t) Factor out e^(-0.2t): C'(t) = e^(-0.2t) * (80 - 16t) --- **Step 2: Set derivative equal to 0 to find critical points** e^(-0.2t) is never 0, so we solve: 80 - 16t = 0 16t = 80 t = 5 hours. --- **Step 3: Verify it's a maximum** We can check the sign of C'(t) around t = 5: For t < 5, say t = 4: 80 - 16*4 = 16 > 0 → C'(t) > 0 → increasing. For t > 5, say t = 6: 80 - 16*6 = -16 < 0 → C'(t) < 0 → decreasing. So t = 5 is indeed a maximum. --- **Step 4: Compute maximum concentration** C(5) = 80 * 5 * e^(-0.2 * 5) C(5) = 400 * e^(-1) C(5) = 400 / e mg/L --- **Final Answer:** Time of maximum concentration: t = 5 hours Maximum concentration: C = 400/e mg/L

6. Sequence: 2, 7, 12, 17... Find a₁₂ and sum of first 12 terms. Answer: a₁₂ = 57, S₁₂ = 354 Solution: Identify the first term a₁ = 2. Find the common difference d = 7 - 2 = 5. Use the nth term formula aₙ = a₁ + (n-1)d.
   Full step-by-step solution

   Step 1: Identify the first term a₁ = 2. Step 2: Find the common difference d = 7 - 2 = 5. Step 3: Use the nth term formula aₙ = a₁ + (n-1)d. For n = 12: a₁₂ = 2 + (12-1)×5 = 2 + 11×5 = 2 + 55 = 57. Step 4: Use the sum formula Sₙ = n(a₁ + aₙ)/2. For n = 12: S₁₂ = 12(2 + 57)/2 = 12×59/2 = 708/2 = 354. Therefore, a₁₂ = 57 and S₁₂ = 354.