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.

Learn how to calculate the 12th term of a geometric sequence where the first term is a_1 = 6 and the common ratio is r = 1/3. This step-by-step solution covers the geometric sequence formula and exponentiation. Suitable for students in Grades 9-11.

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