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The problem states:

"The population of a town three years ago was 'b' and the population of the town three years from now will be 'c'. What is the current population of the town, if it grows at the same rate?"

Let's denote the population now as \( P \).

Given that the population grows at the same rate, we can use the formula for exponential growth or decay:

\[
P = \sqrt{bc}
\]

This means the current population \( P \) is the geometric mean of the population three years ago and three years from now.

From the options given:

1. \( \sqrt{bc} \)
2. \( b \sqrt{c} \)
3. \( c \sqrt{b} \)
4. \( \frac{b \sqrt{b}}{\sqrt{c}} \)

The correct answer is **\( \sqrt{bc} \)**.

Would you like more details or have any questions?

### Related Questions:
1. How do you derive the formula for exponential growth/decay?
2. What is the difference between arithmetic and geometric mean?
3. Can you apply this growth formula to predict populations over longer periods?
4. How does the formula change if growth rates differ over two periods?
5. What are some real-world scenarios where this formula is used?

**Tip:** Always verify your units when dealing with exponential growth in practical applications, such as population studies.

The population of a town three years ago was 'b' and the population of the town three years from now will be 'c'. What is the current population of the town, if it grows at the same rate?

Solve the problem of finding the current population of a town that grows at the same rate. The population three years ago was 'b', and three years from now it will be 'c'. Learn how to apply the geometric mean formula to calculate the current population. Suitable for students in Grades 10-12.

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