Tom started out paying $200, namely the first term in that "geometric sequence" is $200.
and the next payment will be 1.2 times more than the previous, so, if the previous one was 200, the next one will be 200*1.2, and the next after that (200*1.2) * 1.2 and so on.
in a geometric sequence, to get the next term, we simply use a "multiplier", namely the "common ratio", in this case that'd be 1.2.
sigma notation wise,
[tex]\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
n=30\\
a_1=200\\
r=1.2
\end{cases}
\\\\\\
\sum\limits_{i=1}^{30}~200(1.2)^{i-1}[/tex]
and its sum will just be
[tex]\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
n=30\\
a_1=200\\
r=1.2
\end{cases}[/tex]
[tex]\bf S_{30}=200\left( \cfrac{1-1.2^{30}}{1-1.2} \right)\implies S_{30}=200\left( \cfrac{1-\stackrel{\approx}{237.37631}}{-0.2} \right)
\\\\\\
S_{30}=200\left(\cfrac{\stackrel{\approx}{-236.37631}}{-0.2} \right)\implies S_{30}=200(1186.88)
\\\\\\
S_{30}\approx 237366.3137998[/tex]
in plain and short, when the "common ratio" is a fraction, namely less than 1 and less than 0, the serie is convergent, namely it approaches a certain fixed amount.
in this case the common ratio is 1.2, and so is not 0 < | r | < 1, so the serie is divergent, namely it keeps on going.
