Solution:
Since $S_{n}=a_{1} \cdot \frac{1-q^{n}}{1-q}$, the sequence $\left\{S_{n}\right\}_{n=1}^{\infty}$ converges if and only if $|q|<1$. Therefore $-1<\frac{3-2 a}{a-2}<1$, whence $a \in\left(1, \frac{5}{3}\right) \backslash\left\{\frac{3}{2}\right\}$. In this case
$$S=\lim _{n \rightarrow \infty} S_{n}=a_{1} \cdot \frac{1}{1-q}=\frac{3-2 a}{1-\frac{3-2 a}{a-2}}=\frac{(3-2 a)(a-2)}{3 a-5}$$
and we have to prove that for every $a \in\left(1, \frac{5}{3}\right) \backslash\left\{\frac{3}{2}\right\}$ the inequality
$$\frac{(3-2 a)(a-2)}{3 a-5}<1$$
holds. This inequality is equivalent to $\frac{2 a^{2}-4 a+1}{3 a-5}>0$ and since $3 a-5<0$, we have to prove that $f(a)=2 a^{2}-4 a+1<0$. This follows from $f(1)=-1$ and $f\left(\frac{5}{3}\right)=-\frac{1}{9}$.