Let's solve each part **step by step** with explanations after each step, and give the final answer at the end. --- ### (a) Find the sum of the series representing the total fish population over the 24-month period. #### **Step 1: Identify the Series** - The initial population is **200**. - Growth rate is **10% per month**: so each month, population is multiplied by **1.1**. - The population each month forms a geometric sequence: Month 1: 200 Month 2: 200 × 1.1 Month 3: 200 × (1.1)² … Month n: 200 × (1.1)^(n-1) **Series for 24 months:** 200 + 200(1.1) + 200(1.1)² + ... + 200(1.1)²³ **Explanation:** This is a geometric series with first term \( a = 200 \), common ratio \( r = 1.1 \), and number of terms \( n = 24 \). --- #### **Step 2: Use the Sum Formula** The sum of the first \( n \) terms of a geometric series is: \[ S_n = a \frac{1-r^n}{1-r} \] Plug in the values: - \( a = 200 \) - \( r = 1.1 \) - \( n = 24 \) \[ S_{24} = 200 \frac{1 - (1.1)^{24}}{1 - 1.1} \] \[ S_{24} = 200 \frac{1 - (1.1)^{24}}{-0.1} \] \[ S_{24} = -2000 [1 - (1.1)^{24}] \] \[ S_{24} = 2000 [(1.1)^{24} - 1] \] **Final Answer for (a):** \[ \boxed{S_{24} = 2000 \left[(1.1)^{24} - 1\right]} \] This is the sum of the series representing the total fish population over the 24-month period. --- ### (b) Long-term behavior if growth continues indefinitely #### **Step 1: Consider the Formula as \( n \to \infty \)** As \( n \to \infty \), \( (1.1)^n \to \infty \). The sum formula for a geometric series diverges (the sum grows without bound) if \( r > 1 \). **Explanation:** With a constant 10% monthly growth, the population would grow **exponentially** and become infinitely large in the long run, which is not realistic in nature. --- **Final Answer for (b):** If the growth rate continues indefinitely, the fish population will increase without bound (exponentially), eventually becoming infinitely large. --- ### (c) Implications for the lake’s ecosystem #### **Step 1: Analyze Ecological Factors** - **Food supply:** Rapid population growth will increase demand for food, potentially leading to shortages. - **Space:** Overcrowding may occur, causing stress and disease. - **Predation:** Predators may increase due to more prey, but could also be outpaced by fish growth. - **Competition:** Other species may suffer due to increased competition for resources. **Explanation:** Uncontrolled exponential growth can disrupt the ecosystem, leading to resource depletion, imbalance, and possible collapse of the fish population or other species. --- **Final Answer for (c):** Exponential fish population growth can lead to resource depletion, overcrowding, increased disease, and competition, destabilizing the entire ecosystem. --- ### (d) New sequence if growth rate changes to 5% per month after month 12 #### **Step 1: Find Population at Month 12** Population at month 12: \[ P_{12} = 200 \times (1.1)^{11} \] #### **Step 2: Write New Sequence for Months 13 to 24** From month 13 onwards, growth rate = 5% per month (\( r = 1.05 \)). - Month 13: \( P_{12} \times 1.05 \) - Month 14: \( P_{12} \times (1.05)^2 \) - ... - Month 24: \( P_{12} \times (1.05)^{12} \) So, for month \( k \) (where \( k = 1 \) to \( 12 \)), population is: \[ P_{12} \times (1.05)^k = 200 \times (1.1)^{11} \times (1.05)^k \] for \( k = 1, 2, ..., 12 \). **Explanation:** After month 12, the population continues to grow, but at a slower (5%) rate, starting from the value reached at month 12. --- **Final Answer for (d):** \[ P = 200 \times (1.1)^{11} \times (1.05)^k \quad \text{for} \quad k = 1, 2, ..., 12 \ (\text{months 13 to 24}) \] --- Let me know if you need any further clarifications or calculations!

**Final summarized answers:** **(a)** The total fish population over 24 months is: \[ \boxed{2000 \left[(1.1)^{24} - 1\right]} \] **(b)** If growth continues indefinitely, the population grows exponentially without bound (it becomes infinitely large). **(c)** The implications of such exponential growth include resource depletion, overcrowding, increased disease, and ecosystem imbalance. **(d)** After month 12, if the growth rate drops to 5% per month, the population for months 13 to 24 is given by: \[ P = 200 \times (1.1)^{11} \times (1.05)^k, \quad k=1,2,\ldots,12 \]