Find the first term, $a_{1}$, of the geometric series given $S_{7}=6558, r=3$, and $n=7$. Remember to use the formula $S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$. $a_{1}=18$

Step 1 ：Given that the sum of the first 7 terms of the geometric series, \(S_{7}\), is 6558, the common ratio, \(r\), is 3, and the number of terms, \(n\), is 7. We are asked to find the first term, \(a_{1}\).

Step 2 ：We can use the formula for the sum of the first \(n\) terms of a geometric series, which is \(S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}\).

Step 3 ：Substituting the given values into the formula, we get: \(6558 = \frac{a_{1}\left(1-3^{7}\right)}{1-3}\).

Step 4 ：Solving for \(a_{1}\), we first simplify the right side of the equation: \(6558 = \frac{a_{1}\left(1-2187\right)}{-2}\).

Step 5 ：Simplify the equation to: \(6558 = \frac{-2186a_{1}}{-2}\).

Step 6 ：Further simplify the equation to: \(6558 = 1093a_{1}\).

Step 7 ：Finally, we divide both sides of the equation by 1093 to solve for \(a_{1}\): \(a_{1} = \frac{6558}{1093}\).

Step 8 ：Simplify to get the final answer: \(\boxed{a_{1} = 6}\).