We are tasked with solving the given geometric sequence problem based on the provided formula:

$a_n = a_1 \cdot q^{n-1}$

Problem Details:

• $a_1 = 1$ (initial number of cells)
• $q = 3$ (growth rate)
• $n = 4$ (time step we are interested in)

We need to find $a_4$, the number of cells at time $n = 4$.

Solution:

Using the formula: $a_4 = a_1 \cdot q^{4-1}$ Substitute the values: $a_4 = 1 \cdot 3^{3}$ Simplify the exponentiation: $3^3 = 3 \cdot 3 \cdot 3 = 27$ Thus: $a_4 = 1 \cdot 27 = 27$

Final Answer:

The number of cells at time $a_4$ is 27.

Would you like a detailed explanation of the concept of geometric sequences, or have any related questions?

Here are some related questions you might explore:

1. What happens to the growth rate if $q < 1$?
2. How does the formula change if we start indexing at $n = 0$?
3. Can this formula model population decline? Why or why not?
4. How does increasing $q$ impact the sequence over time?
5. If $a_1 = 2$, how would that impact the result at $a_4$?

Tip: Always check if the initial value ($a_1$) and growth rate ($q$) are reasonable for the context you're modeling, as extreme values can skew results!