To describe the situation mathematically, we can represent the total population of the animals as a geometric sequence. The initial population \( P_0 \) is 4000, and the population decreases by 40% each year. This means that each year, 60% of the population remains.
The first term (\( P_0 \)) of the sequence is: \[ P_0 = 4000 \]
The common ratio (\( r \)) of the sequence, which represents the population remaining after each year, is: \[ r = 1 - 0.40 = 0.60 \]
The sequence can be expressed as: \[ P_n = P_0 \cdot r^n \] where:
- \( P_n \) is the population in year \( n \),
- \( n \) is the number of years.
Now, let's calculate the first 10 terms of the sequence:
- \( P_0 = 4000 \)
- \( P_1 = 4000 \cdot 0.60^1 = 4000 \cdot 0.60 = 2400 \)
- \( P_2 = 4000 \cdot 0.60^2 = 4000 \cdot 0.36 = 1440 \)
- \( P_3 = 4000 \cdot 0.60^3 = 4000 \cdot 0.216 = 864 \)
- \( P_4 = 4000 \cdot 0.60^4 = 4000 \cdot 0.1296 = 518.4 \)
- \( P_5 = 4000 \cdot 0.60^5 = 4000 \cdot 0.07776 = 310.32 \)
- \( P_6 = 4000 \cdot 0.60^6 = 4000 \cdot 0.046656 = 186.624 \)
- \( P_7 = 4000 \cdot 0.60^7 = 4000 \cdot 0.0279936 = 111.9744 \)
- \( P_8 = 4000 \cdot 0.60^8 = 4000 \cdot 0.01679616 = 67.18464 \)
- \( P_9 = 4000 \cdot 0.60^9 = 4000 \cdot 0.010077696 = 40.310784 \)
So the first 10 terms of the sequence representing the animal population over the years are approximately:
- \( P_0 \approx 4000 \)
- \( P_1 \approx 2400 \)
- \( P_2 \approx 1440 \)
- \( P_3 \approx 864 \)
- \( P_4 \approx 518.4 \)
- \( P_5 \approx 310.32 \)
- \( P_6 \approx 186.624 \)
- \( P_7 \approx 111.9744 \)
- \( P_8 \approx 67.18464 \)
- \( P_9 \approx 40.310784 \)
This sequence reflects the decreasing population of the animals over 10 years, showing how significantly the population declines as a result of the 40% annual decrease.
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