The population of a town in 2010 is 5000. The population of the town for the next five successive
years is recorded as: 5421, 5836,
6293, 6725, 7202. Assuming the population is approximately
following an arithmetic sequence, find (to an accuracy of three significant figures) the common difference of the sequence and a prediction for the town's population in 2024.

We are given the population data for a town that seems to follow an arithmetic sequence:

- Population in 2010: 5000
- Population for the next five successive years: 5421, 5836, 6293, 6725, 7202

### Step 1: Verify Arithmetic Sequence

In an arithmetic sequence, the difference between consecutive terms (called the **common difference, \( d \)**) remains constant. We can check this by calculating the differences between the successive populations:

- \( 5421 - 5000 = 421 \)
- \( 5836 - 5421 = 415 \)
- \( 6293 - 5836 = 457 \)
- \( 6725 - 6293 = 432 \)
- \( 7202 - 6725 = 477 \)

While the differences are not exactly the same, they are close enough, suggesting an approximate arithmetic sequence.

### Step 2: Calculate the Average Common Difference

Since the differences vary slightly, we will take the **average** of these differences to find an approximate common difference:

\[
d = \frac{421 + 415 + 457 + 432 + 477}{5} = \frac{2202}{5} = 440.4
\]

Thus, the approximate common difference is \( d = 440.4 \).

### Step 3: Predict Population in 2024

The year 2024 is 14 years after 2010, so we need to find the population after 14 terms. Using the formula for the \( n \)-th term of an arithmetic sequence:

\[
a_n = a_1 + (n-1)d
\]

Where:
- \( a_1 = 5000 \) (the population in 2010),
- \( d = 440.4 \) (the common difference),
- \( n = 15 \) (since 2024 is the 15th year after 2010).

Now, we calculate \( a_{15} \):

\[
a_{15} = 5000 + (15-1) \times 440.4 = 5000 + 14 \times 440.4 = 5000 + 6165.6 = 11165.6
\]

So, the predicted population in 2024 is approximately **11,200** (to three significant figures).

### Summary:

- **Common difference**: \( 440.4 \)
- **Predicted population in 2024**: \( 11,200 \)

Do you have any questions, or would you like more details?

---

Here are 5 related questions for further practice:
1. How would the prediction change if the common difference were slightly larger or smaller?
2. What is the population prediction for 2030 assuming the same arithmetic pattern?
3. Can you model this population growth using a geometric sequence instead?
4. What would be the effect of a decreasing common difference on the population growth?
5. How can we account for external factors like migration or birth rate changes in this model?

**Tip:** Arithmetic sequences are useful for modeling consistent growth patterns, but remember that real-world factors can cause fluctuations, making precise predictions difficult.

The population of a town in 2010 is 5000. The population of the town for the next five successive years is recorded as: 5421, 5836, 6293, 6725, 7202. Assuming the population is approximately following an arithmetic sequence, find (to an accuracy of three significant figures) the common difference of the sequence and a prediction for the town's population in 2024.

Learn how to predict the population of a town using an arithmetic sequence. Starting with a population of 5000 in 2010, the next five years' populations are recorded as 5421, 5836, 6293, 6725, and 7202. Calculate the common difference and predict the population for 2024, suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation