The sum of the first \(n\) terms of a series is given by the expression
\[\begin{equation*}
6 - \frac{2^{n+1}}{3^{n-1}}.
\end{equation*}\]

Let the sequence that we are summing be \((a_n)\) for \(n \ge 1\), so that
\[\begin{equation*}
a_1 + \dotsb + a_n = 6 - \frac{2^{n+1}}{3^{n-1}}
\end{equation*}\]
for every \(n \ge 1\). In particular,
\[\begin{equation*}
a_1 = 6 - \frac{2^2}{3^0} = 6 - 4 = 2,
\end{equation*}\]
and, for every \(n \ge 2\),
\[\begin{align*}
a_{n} = \left( a_1 + \dotsb + a_{n} \right) - \left( a_1 + \dotsb + a_{n-1} \right) &= \left( 6 - \frac{2^{n+1}}{3^{n-1}} \right) - \left( 6 - \frac{2^{n}}{3^{n-2}} \right) \\
&= \frac{2^{n}}{3^{n-2}} - \frac{2^{n+1}}{3^{n-1}} \\
&= \frac{2^{n}}{3^{n-2}} \left( 1 - \frac{2}{3} \right) \\
&= \frac{2^{n}}{3^{n-2}} \frac{1}{3} \\
&= \frac{2^{n}}{3^{n-1}} \\
&= 2 \left( \frac{2}{3} \right)^{n-1}.
\end{align*}\]
By finding an expression for the \(n\)th term of the series, or otherwise, show that this is a geometric series, and state the values of the first term and the common ratio.

Thus, \((a_n)\) is a geometric progression with first term \(a_1 = 2\) and common ratio \(\dfrac{2}{3}\).