which term of the following geometric sequence 1,3,9... has the value of 19683

In a geometric sequence, the \(n\)-th term is given by:

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

where:
- \(T_n\) is the \(n\)-th term,
- \(a\) is the first term,
- \(r\) is the common ratio,
- \(n\) is the term number.

For the given sequence \(1, 3, 9, \dots\), we can identify:
- \(a = 1\) (first term),
- \(r = 3\) (common ratio, because \( \frac{3}{1} = \frac{9}{3} = 3 \)).

Now, we need to find \(n\) for which \(T_n = 19683\).

Using the formula for the \(n\)-th term:

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

Set \(T_n = 19683\):

\[
19683 = 3^{n-1}
\]

To solve for \(n\), we express \(19683\) as a power of 3. We know:

\[
19683 = 3^9
\]

Thus, we have:

\[
3^{n-1} = 3^9
\]

Equating the exponents:

\[
n - 1 = 9
\]

So:

\[
n = 10
\]

Therefore, the 10th term of the sequence is \(19683\).

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Would you like more details or have any questions?

Here are some related questions for further practice:

1. What is the 7th term of the sequence \(1, 3, 9, 27, \dots\)?
2. How do you find the sum of the first 10 terms of this geometric sequence?
3. What is the 15th term in the geometric sequence \(2, 6, 18, \dots\)?
4. What is the formula for the sum of an infinite geometric series?
5. How can you find the common ratio if only two non-consecutive terms are known?

**Tip:** When solving geometric sequence problems, recognizing powers of the common ratio helps simplify the process of finding specific terms.

Which term of the following geometric sequence 1, 3, 9... has the value of 19683?

Learn how to find which term of the geometric sequence 1, 3, 9... has the value of 19683. This problem involves key concepts from geometric sequences and exponentiation, and is suitable for students in Grades 9-10.

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