Infinite Geometric Series

lessonbunny.com

Answer Key & Explanations

Infinite Geometric Series · Grade 12 · Worksheet 3

1. Mason is studying the long-term accumulation of a new slow-release fertilizer in a garden plot. Each week, the soil absorbs 72 mg of the fertilizer from the latest application, but 30% of the existing fertilizer is consumed by plants or leached away by the end of each week. If Mason applies 72 mg of fertilizer every week indefinitely, what total amount of fertilizer (in mg) will be present in the soil in the long run? Answer: 240 Solution: Identify the pattern. Each week, Mason adds 72 mg. At the end of the week, 30% of the existing fertilizer is lost, meaning 70% remains.
   Full step-by-step solution

   Step 1: Identify the pattern. Each week, Mason adds 72 mg. At the end of the week, 30% of the existing fertilizer is lost, meaning 70% remains. So the fertilizer from each week's application decays geometrically: the first week's contribution is 72 mg, the second week's contribution is 72 * 0.7, the third week's is 72 * (0.7)^2, and so on. Step 2: This is an infinite geometric series with first term a = 72 and common ratio r = 0.7. Step 3: Check convergence. Since |r| = 0.7 < 1, the series converges. Step 4: Apply the infinite geometric series sum formula: S = a / (1 - r). Step 5: Calculate: S = 72 / (1 - 0.7) = 72 / 0.3 = 240. Step 6: The total amount of fertilizer that will be present in the soil in the long run is 240 mg. The answer is 240.

2. Olivia is monitoring the spread of a beneficial bacteria in a bioreactor used for waste treatment. Each hour, the bacteria break down 80% of the remaining waste in the reactor, leaving 20% of the waste untouched. If the initial waste load is 500 kg, and the bacteria continue this pattern indefinitely, what is the total amount of waste (in kg) that the bacteria will eventually break down? Answer: 500 Solution: Determine the first term (waste broken down in the first hour). The initial waste is 500 kg. The bacteria break down 80% of the waste each hour, so in the first hour, they break down 0.80 * 500 = 400 kg.
   Full step-by-step solution

   Step 1: Determine the first term (waste broken down in the first hour). The initial waste is 500 kg. The bacteria break down 80% of the waste each hour, so in the first hour, they break down 0.80 * 500 = 400 kg. Step 2: After the first hour, 500 - 400 = 100 kg remains. In the second hour, they break down 80% of the remaining 100 kg, which is 0.80 * 100 = 80 kg. Step 3: The waste broken down each hour forms a geometric sequence: 400, 80, 16, ... The common ratio r = 80 / 400 = 0.20 (since each hour they break down 20% of the previous hour's breakdown). Step 4: Since |r| = 0.20 < 1, the infinite geometric series converges. The sum S = a / (1 - r), where a = 400 and r = 0.20. Step 5: Calculate S = 400 / (1 - 0.20) = 400 / 0.80 = 500. Step 6: The total amount of waste the bacteria will eventually break down is 500 kg. The answer is 500.

3. A pharmaceutical company is testing a new drug that is eliminated from the body at a rate such that 60% of the drug remains in the bloodstream at the end of each 8-hour dosing interval. If a patient takes a 250 mg dose every 8 hours, what is the maximum possible concentration of the drug in the bloodstream that will be approached over time? Answer: 625 Solution: Let’s go step-by-step. The patient takes 250 mg every 8 hours. At the end of each 8-hour period, 60% of the drug remains in the bloodstream.
   Full step-by-step solution

   Let’s go step-by-step. --- **Step 1: Understand the problem** The patient takes 250 mg every 8 hours. At the end of each 8-hour period, 60% of the drug remains in the bloodstream. That means 40% is eliminated over each 8-hour interval. We want the maximum possible concentration approached over time — this is the **steady-state level** just after a dose. --- **Step 2: Set up the sequence** Let \( M_n \) be the amount of drug in the bloodstream **right after** the \( n \)-th dose. Initially: \( M_1 = 250 \) mg (first dose) Right before the 2nd dose: 60% of \( M_1 \) remains: \( 0.60 \times 250 = 150 \) mg Then the 2nd dose of 250 mg is added: \( M_2 = 150 + 250 = 400 \) mg Right before the 3rd dose: 60% of \( M_2 \) remains: \( 0.60 \times 400 = 240 \) mg Then the 3rd dose of 250 mg is added: \( M_3 = 240 + 250 = 490 \) mg We see a pattern: \( M_{n} = 250 + 0.60 \times M_{n-1} \) --- **Step 3: Find the steady-state level** At steady state, \( M_n = M_{n-1} = M \) (just after dose). From the pattern: \( M = 250 + 0.60 \times M \) Solve for \( M \): \( M - 0.60M = 250 \) \( 0.40M = 250 \) \( M = 250 / 0.40 \) \( M = 625 \) --- **Step 4: Interpret the result** The maximum possible concentration approached over time is **625 mg** in the bloodstream (right after a dose). --- **Final answer:** 625

4. A wildlife biologist is studying the spread of a rare bird species in a protected forest. Each breeding pair produces offspring that establish new territories, but due to limited habitat, only 30% of these new pairs successfully reproduce in the next generation. If the forest currently has 80 breeding pairs, and this pattern continues indefinitely, what is the theoretical maximum total number of breeding pairs the forest could eventually support? Answer: 114 Solution: Identify the initial number of breeding pairs: 80 Identify the common ratio: 30% = 0.3 This forms an infinite geometric series: 80 + 80(0.3) + 80(0.3)^2 + 80(0.3)^3 + ...
   Full step-by-step solution

   Step 1: Identify the initial number of breeding pairs: 80 Step 2: Identify the common ratio: 30% = 0.3 Step 3: This forms an infinite geometric series: 80 + 80(0.3) + 80(0.3)^2 + 80(0.3)^3 + ... Step 4: Use the infinite geometric series formula: S = a / (1 - r), where a = 80 and r = 0.3 Step 5: Calculate: S = 80 / (1 - 0.3) = 80 / 0.7 Step 6: 80 ÷ 0.7 = 114.2857... Step 7: Since we're counting breeding pairs, we round to the nearest whole number: 114 The answer is 114.

5. ∑(n=1 to ∞) 5(0.8)^(n-1) = ? Answer: 25 Solution: The series is 5(0.8)^(n-1), so a = 5 and r = 0.8 Since |r| = |0.8| = 0.8 < 1, the series converges Sum = a / (1 - r) Sum = 5 / (1 - 0.8) 1 - 0.8 = 0.2 5 / 0.2 = 25 The answer is 25.
   Full step-by-step solution

   Step 1: Identify the first term (a) and common ratio (r) The series is 5(0.8)^(n-1), so a = 5 and r = 0.8 Step 2: Check convergence condition Since |r| = |0.8| = 0.8 < 1, the series converges Step 3: Apply the infinite geometric series formula Sum = a / (1 - r) Sum = 5 / (1 - 0.8) Step 4: Calculate the denominator 1 - 0.8 = 0.2 Step 5: Divide to find the sum 5 / 0.2 = 25 The answer is 25.

6. Kaia is a seismologist analyzing the energy released by a series of aftershocks following a major earthquake. The first aftershock releases 9,000 megajoules of energy. Each subsequent aftershock releases 80% of the energy of the previous one. If this pattern of aftershocks continues indefinitely, what is the total energy released by all the aftershocks? Answer: 45000 megajoules Solution: Identify the first term and common ratio. The first term a = 9000 megajoules. Each subsequent term is 80% of the previous, so the common ratio r = 0.80.
   Full step-by-step solution

   Step 1: Identify the first term and common ratio. The first term a = 9000 megajoules. Each subsequent term is 80% of the previous, so the common ratio r = 0.80. Step 2: Check convergence. Since |r| = 0.80 < 1, the infinite geometric series converges. Step 3: Apply the infinite geometric series sum formula: S = a / (1 - r). Step 4: Substitute the values: S = 9000 / (1 - 0.80) = 9000 / 0.20. Step 5: Simplify: 9000 / 0.20 = 9000 / (1/5) = 9000 * 5 = 45000. Step 6: The total energy released by all aftershocks is 45000 megajoules.