The problem describes a bacterial population growing according to a geometric sequence, with given populations on specific days.

We know:

• Day 4: 4,000 bacteria
• Day 8: 49,000 bacteria
• The growth follows a geometric sequence.

To solve for the population on day 6, we need to:

1. Use the geometric sequence formula $a_n = a_1 \cdot r^{n-1}$, where:
   • $a_n$ is the population on day $n$,
   • $a_1$ is the initial population,
   • $r$ is the common ratio.

Step 1: Set up the relationship for day 4 and day 8

Let $a_4 = 4,000$ and $a_8 = 49,000$. This means: $a_8 = a_4 \cdot r^4$ $49,000 = 4,000 \cdot r^4$ Solving for $r$: $r^4 = \frac{49,000}{4,000} = 12.25$ $r = \sqrt[4]{12.25} = 1.8$

Step 2: Calculate the population on day 6

Now that we know $r = 1.8$, we can calculate the population on day 6 using: $a_6 = a_4 \cdot r^2$ $a_6 = 4,000 \cdot (1.8)^2 = 4,000 \cdot 3.24 = 12,960$

Since none of the options matches exactly, the closest option is 14,000 (Option C), which likely rounds the calculated answer.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you find the common ratio in a geometric sequence?
2. What is the formula for the general term of a geometric sequence?
3. How does rounding affect answers in real-world problems?
4. Why is it important to check units in word problems?
5. What is the difference between arithmetic and geometric sequences?

Tip: Always recheck rounding when solving problems with real-world contexts like population growth.