In a geometric progression (GP), each term is obtained by multiplying the previous term by a constant ratio. The general formula for the \(n\)-th term of a GP is:

\[
T_n = a \cdot r^{n-1}
\]

Where:
- \(T_n\) is the \(n\)-th term,
- \(a\) is the first term,
- \(r\) is the common ratio,
- \(n\) is the position of the term.

For the given GP: 3, 9, 27, ..., we know:
- \(a = 3\),
- The common ratio \(r = \frac{9}{3} = 3\),
- The term we are looking for is 729.

We need to find \(n\) such that:

\[
T_n = 3 \cdot 3^{n-1} = 729
\]

Dividing both sides by 3:

\[
3^{n-1} = \frac{729}{3} = 243
\]

Now, express 243 as a power of 3:

\[
243 = 3^5
\]

So:

\[
3^{n-1} = 3^5
\]

This implies:

\[
n - 1 = 5 \quad \Rightarrow \quad n = 6
\]

Therefore, the 6th term of the GP is 729.

### Would you like more details or have any questions?

Here are 5 related questions:
1. How can we find the sum of a certain number of terms in a GP?
2. How do we determine if a sequence is geometric?
3. What is the formula for the sum of an infinite GP?
4. How do arithmetic progressions (AP) differ from geometric progressions (GP)?
5. Can we derive the common ratio if only a few terms of a GP are given?

**Tip**: When working with powers and logarithms in GPs, knowing how to express numbers as powers of a base can simplify calculations significantly.

Discover how to find the term of a geometric progression, specifically which term of the sequence 3, 9, 27 equals 729. This problem covers key concepts of geometric progressions, exponentiation, and includes a step-by-step solution. Ideal for students in Grades 9-11.

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