You are dealing with a geometric sequence, where the first term $a_1 = 6$ and the common ratio $r = \frac{1}{3}$. We need to find the 12th term $a_{12}$.

The general formula for the $n$-th term of a geometric sequence is:

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

Substitute the given values into the formula:

$a_{12} = 6 \cdot \left(\frac{1}{3}\right)^{12-1} = 6 \cdot \left(\frac{1}{3}\right)^{11}$

Now let's compute the value of $a_{12}$:

$a_{12} = 6 \cdot \frac{1}{3^{11}} = 6 \cdot \frac{1}{177147}$

Simplifying:

$a_{12} = \frac{6}{177147}$

Thus, the 12th term is:

$a_{12} = \frac{2}{59049}$

Would you like further clarification or details on this process?

Here are some related questions:

1. How would the 12th term change if $r = \frac{1}{2}$?
2. What is the sum of the first 12 terms of this geometric series?
3. How does a geometric sequence differ from an arithmetic sequence?
4. What happens to the terms as $n$ approaches infinity in a geometric sequence with $|r| < 1$?
5. How would the 12th term formula change if we were dealing with an arithmetic sequence?

Tip: For geometric sequences, if the common ratio $|r| < 1$, the terms get progressively smaller as $n$ increases.