Find the common ratio and write out the first four terms of the geometric sequence $\frac{2^{n-3}}{6}$

Common ratio is
\[
a_{1}=\square, a_{2}=\square, a_{3}=\square, a_{4}=\square
\]

Question Help: $\square$ Message instructor
Submit Question

Step 1 ：The common ratio of a geometric sequence is the ratio between any term and the previous term. In this case, the common ratio is 2, because each term is multiplied by 2 to get the next term.

Step 2 ：The first four terms can be found by substituting n=1, n=2, n=3, and n=4 into the formula for the nth term.

Step 3 ：Let's calculate these values. For n=1, \(a_{1} = \frac{2^{1-3}}{6} = \frac{1}{24}\). For n=2, \(a_{2} = \frac{2^{2-3}}{6} = \frac{1}{12}\). For n=3, \(a_{3} = \frac{2^{3-3}}{6} = \frac{1}{6}\). For n=4, \(a_{4} = \frac{2^{4-3}}{6} = \frac{1}{3}\).

Step 4 ：\(\boxed{\text{The common ratio is } 2 \text{ and the first four terms of the geometric sequence are } \frac{1}{24}, \frac{1}{12}, \frac{1}{6}, \text{ and } \frac{1}{3}}\)