Find the 10th term of the sequence 2/3, 4/9, 8/27....

The given sequence is:

\[
\frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \dots
\]

This is a geometric sequence where the first term \( a = \frac{2}{3} \) and the common ratio \( r \) is:

\[
r = \frac{\frac{4}{9}}{\frac{2}{3}} = \frac{4}{9} \times \frac{3}{2} = \frac{2}{3}
\]

The general formula for the \(n\)-th term of a geometric sequence is:

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

Substituting the known values to find the 10th term:

\[
a_{10} = \frac{2}{3} \cdot \left( \frac{2}{3} \right)^{10-1} = \frac{2}{3} \cdot \left( \frac{2}{3} \right)^9
\]

Now let's compute:

\[
a_{10} = \frac{2}{3} \cdot \frac{2^9}{3^9} = \frac{2^{10}}{3^{10}}
\]

\[
a_{10} = \frac{1024}{59049}
\]

Thus, the 10th term of the sequence is:

\[
\frac{1024}{59049}
\]

Would you like more details on any part of this solution?

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Here are 5 related questions:
1. How do you find the sum of the first \(n\) terms of a geometric sequence?
2. What is the common ratio in a geometric sequence?
3. How do you determine if a sequence is arithmetic or geometric?
4. Can geometric sequences have negative common ratios?
5. How do you find the limit of a geometric sequence as \(n \to \infty\)?

**Tip:** Always check for the common ratio by dividing consecutive terms when determining if a sequence is geometric.

Discover how to find the 10th term of the geometric sequence 2/3, 4/9, 8/27 using the nth term formula. This problem highlights important algebraic concepts and is suitable for students in Grades 9-12.

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