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To solve this problem of finding the 6th term of a geometric sequence where the common ratio is \( \frac{1}{3} \) and the first term is 5, we use the formula for the \(n\)-th term of a geometric sequence:

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

Where:
- \( a_n \) is the \(n\)-th term,
- \( a_1 \) is the first term (in this case, 5),
- \( r \) is the common ratio (here, \( \frac{1}{3} \)),
- \( n \) is the term number (6, in this case).

Substituting the values:

\[
a_6 = 5 \cdot \left( \frac{1}{3} \right)^{6-1} = 5 \cdot \left( \frac{1}{3} \right)^5
\]

Now, let's calculate that.The 6th term of the geometric sequence is approximately \( 0.0206 \).

Would you like further details or have any other questions? Here are some related questions for you to explore:

1. How do you find the sum of the first 6 terms of this geometric series?
2. What is the 10th term of the same geometric sequence?
3. How does the geometric series behave as \(n\) approaches infinity with this common ratio?
4. Can you derive the general formula for the sum of an infinite geometric series?
5. What happens to the terms of the sequence if the common ratio is changed to a value greater than 1?

**Tip**: In a geometric sequence, if the common ratio is less than 1, the terms decrease exponentially.

Hallar el 6.° término de la sucesión geométrica cuya razón común es 1/3 y cuyo primer término es 5.

Find the 6th term of a geometric sequence with a common ratio of 1/3 and a first term of 5. The solution involves using the formula for the nth term of a geometric sequence and performing exponentiation.

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