The given geometric sequence is:

$2, \frac{2}{5}, \frac{2}{25}, \frac{2}{125}, \dots$

To find $a_9$, the 9th term of the sequence, we need to first identify the common ratio $r$ and then use the formula for the $n$-th term of a geometric sequence:

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

Step 1: Find the common ratio $r$

The common ratio $r$ is found by dividing any term by the previous term. Let's divide the second term by the first term:

$r = \frac{\frac{2}{5}}{2} = \frac{2}{5} \times \frac{1}{2} = \frac{1}{5}$

So, the common ratio $r = \frac{1}{5}$.

Step 2: Use the formula for the $n$-th term

Now, we can use the formula to find $a_9$:

$a_9 = a_1 \cdot r^{9-1} = 2 \cdot \left(\frac{1}{5}\right)^8$

Step 3: Simplify the expression

$a_9 = 2 \cdot \frac{1}{5^8} = 2 \cdot \frac{1}{390625}$

$a_9 = \frac{2}{390625}$

So, the 9th term $a_9 = \frac{2}{390625}$.

Let me know if you need further details or clarifications.

Related Questions:

1. How do you find the common ratio of a geometric sequence?
2. What happens to the terms of a geometric sequence as the common ratio gets smaller?
3. How would you solve for the 10th term of this sequence?
4. What is the sum of the first 9 terms of a geometric sequence?
5. How do you check if a sequence is geometric?

Tip: When working with geometric sequences, the common ratio is key, and once you identify it, you can easily find any term using the formula $a_n = a_1 \cdot r^{n-1}$.