Write a formula for the nth term of the following geometric sequence.
\[
-2, \frac{2}{5},-\frac{2}{25}, \frac{2}{125} \ldots
\]

Find a formula for the nth term of the geometric sequence.
\[
a_{n}=\square \cdot(\square)^{n-1}
\]

Step 1 ：Given a geometric sequence \(-2, \frac{2}{5}, -\frac{2}{25}, \frac{2}{125} \ldots\)

Step 2 ：The first term of the sequence is -2 and the common ratio is -1/5.

Step 3 ：The nth term of a geometric sequence can be found using the formula: \(a_{n}=a_{1} \cdot r^{n-1}\) where \(a_{1}\) is the first term, r is the common ratio, and n is the term number.

Step 4 ：In this case, \(a_{1}=-2\) and \(r=-1/5\).

Step 5 ：Substitute \(a_{1}\) and r into the formula, we get the formula for the nth term of the geometric sequence is \(a_{n}=-2 \cdot\left(-\frac{1}{5}\right)^{n-1}\)

Step 6 ：\(\boxed{a_{n}=-2 \cdot\left(-\frac{1}{5}\right)^{n-1}}\) is the final answer.