The problem involves finding the ninth term ($a_9$) of a geometric sequence where $a_3 = 18$ and $a_6 = 486$.

In a geometric sequence, each term is given by: $a_n = a_1 \cdot r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio.

1. Set up equations using the given terms:

   For $a_3$: $a_3 = a_1 \cdot r^{3-1} = a_1 \cdot r^2 = 18$

   For $a_6$: $a_6 = a_1 \cdot r^{6-1} = a_1 \cdot r^5 = 486$

2. Divide the second equation by the first to solve for $r$: $\frac{a_1 \cdot r^5}{a_1 \cdot r^2} = \frac{486}{18}$ Simplifying, we get: $r^3 = 27$ Taking the cube root of both sides, we find: $r = 3$

3. Substitute $r = 3$ back into the first equation to find $a_1$: $a_1 \cdot 3^2 = 18$ $a_1 \cdot 9 = 18$ $a_1 = 2$

4. Find $a_9$:

   Now, use the formula for the $n$-th term with $n = 9$: $a_9 = a_1 \cdot r^{9-1} = 2 \cdot 3^8$ Calculating $3^8$: $3^8 = 6561$ So, $a_9 = 2 \cdot 6561 = 13122$

Therefore, $a_9 = 13122$.

Would you like more details on any specific step?

Here are five related questions for further exploration:

1. How would you find $a_{10}$ in this sequence?
2. What if $a_1$ and $r$ were different—how would that affect the sequence?
3. Can you calculate the sum of the first 9 terms in this sequence?
4. What is the general formula for the sum of the first $n$ terms of a geometric sequence?
5. How would you determine if a sequence is geometric or arithmetic?

Tip: In a geometric sequence, dividing consecutive terms can quickly reveal the common ratio $r$.