To solve the problem, we use the formula for the $n$-th term of a geometric sequence:

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

Given:

• First term ($a_1$) = 1
• Common ratio ($r$) = $-\frac{1}{3}$
• Term to find ($n$) = 9

Substituting into the formula:

$a_9 = 1 \cdot \left(-\frac{1}{3}\right)^{9-1}$

$a_9 = \left(-\frac{1}{3}\right)^8$

Since raising $-\frac{1}{3}$ to an even power results in a positive value:

$a_9 = \frac{1^8}{3^8} = \frac{1}{6561}$

Final Answer:

The ninth term is:

$a_9 = \frac{1}{6561}$

Would you like further details or clarification? Here are 5 related questions to deepen your understanding:

1. How would the solution change if $n$ were an odd number instead?
2. Why does the sign change in geometric sequences with negative ratios when raising to odd and even powers?
3. How can you express the sequence in explicit terms for general $n$?
4. What is the sum of the first 9 terms of this geometric sequence?
5. How does a geometric sequence differ from an arithmetic sequence?

Tip: Always check the parity (odd or even) of the power in geometric sequences with a negative common ratio—it affects the sign of the result.