The sum, $S_n$, of the first $n$ terms of a geometric progression is given by $S_n = 6 - \dfrac{2}{3^{n-1}}$. Find the first term and the common ratio.

1. Find the first term by evaluating $S_1$.

The first term, $a$, is equal to the sum of the first term, $S_1$.
$a = S_1 = 6 - \dfrac{2}{3^{1-1}}$
$a = 6 - \dfrac{2}{3^0}$
$a = 6 - 2 = 4$

2. Evaluate the sum of the first two terms, $S_2$, and find the second term.

$S_2 = 6 - \dfrac{2}{3^{2-1}}$
$S_2 = 6 - \dfrac{2}{3} = \dfrac{16}{3}$
The second term, $u_2$, is the difference between $S_2$ and $S_1$.
$u_2 = S_2 - S_1$
$u_2 = \dfrac{16}{3} - 4 = \dfrac{4}{3}$

3. Calculate the common ratio using the first and second terms.

$r = \dfrac{u_2}{a}$
$r = \dfrac{4/3}{4} = \dfrac{1}{3}$