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.

To solve this, we need to analyze the population values as an arithmetic sequence.

### Step 1: Verify that the population is in an arithmetic sequence
In an arithmetic sequence, the difference between consecutive terms is constant. Let's calculate the differences between each year:

1. Difference between 2011 and 2010:
\[
5421 - 5000 = 421
\]

2. Difference between 2012 and 2011:
\[
5836 - 5421 = 415
\]

3. Difference between 2013 and 2012:
\[
6293 - 5836 = 457
\]

4. Difference between 2014 and 2013:
\[
6725 - 6293 = 432
\]

5. Difference between 2015 and 2014:
\[
7202 - 6725 = 477
\]

The differences are not exactly the same but are close. We can approximate the common difference by taking the average of these values:

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

### Step 2: Use the arithmetic sequence formula
The formula for the \(n\)-th term of an arithmetic sequence is:

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

Where:
- \(a_n\) is the population in year \(n\),
- \(a_1 = 5000\) (the population in 2010),
- \(d = 440.4\) (the common difference),
- \(n\) is the number of years after 2010.

### Step 3: Predict the population in 2024
To find the population in 2024, we need \(n = 2024 - 2010 = 14\).

Using the formula for the 14th term:
\[
a_{14} = 5000 + (14 - 1) \times 440.4 = 5000 + 13 \times 440.4
\]
\[
a_{14} = 5000 + 5725.2 = 10725.2
\]

Thus, the predicted population in 2024 is approximately **10,700** (to three significant figures).

Would you like more details on this, or do you have any questions?

### Related Questions:
1. How does an arithmetic sequence differ from a geometric sequence?
2. How can we estimate a more accurate common difference using statistical methods?
3. What are the limitations of using arithmetic sequences for population predictions?
4. How would the prediction change if we used a geometric model instead?
5. How does rounding affect predictions in sequences like this?

### Tip:
When working with sequences, always check if the differences or ratios between terms are consistent before assuming the type of progression.

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 in 2024, starting with a population of 5000 in 2010. Using arithmetic sequences, we calculate the common difference and provide a detailed step-by-step solution, suitable for students in Grades 9-11.

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