14. (25 points) If $\left\{a_{n}\right\}$ is an arithmetic sequence, and satisfies $3 a_{5}=8 a_{12}>0$; the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=a_{n} a_{n+1} a_{n+2}$ $\left(n \in \mathbf{Z}_{+}\right)$, and the sum of the first $n$ terms of $\left\{b_{n}\right\}$ is denoted as $S_{n}$. Question: For what value of $n$ does $S_{n}$ achieve its maximum value? Explain your reasoning.
14. From $3 a_{5}=8 a_{12}=8\left(a_{5}+7 d\right)>0$, we know
$$
a_{5}=-\frac{56}{5} d>0 \Rightarrow d<0, a_{17}a_{2}>\cdots>a_{16}>0>a_{17}>a_{18}>\cdots$,
$$
$b_{1}>b_{2}>\cdots>b_{14}>0>b_{17}>b_{18}>\cdots$. 
$$

And $b_{15}=a_{15} a_{16} a_{17}0$, so $S_{14}>S_{13}>\cdots>S_{1}, S_{14}>S_{15}$,
$$
\begin{array}{l}
S_{15}S_{17}>S_{18}>\cdots \\
\text { Also, } S_{16}-S_{14}=b_{15}+b_{16}=a_{16} a_{17}\left(a_{15}+a_{18}\right) \\
=a_{16} a_{17}\left(-\frac{6}{5} d+\frac{9}{5} d\right)=\frac{3}{5} d a_{16} a_{17}>0,
\end{array}
$$

Therefore, $S_{n}$ is maximized at $S_{16}$, i.e., $n=16$ when $S_{n}$ reaches its maximum value.