5. Given an arithmetic sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}<0, a_{100} \geqslant 74, a_{200}<200 \text {, }
$$

and the number of terms of this sequence in the interval $\left(\frac{1}{2}, 8\right)$ is 2 less than the number of terms in the interval $\left[14, \frac{43}{2}\right]$. Then the general term formula of $\left\{a_{n}\right\}$ is
5. $a_{n}=\frac{3}{4} n-1$.

From the problem, we know that $\frac{15}{2}$ and $6$ are both multiples of the common difference $d$. Let $d=\frac{3}{2 m}\left(m \in \mathbf{Z}_{+}\right)$.
From the given conditions, we solve to get
$$
\begin{array}{l}
\frac{74}{99}a_{1} \geqslant 74-99 d=-\frac{1}{4} .
\end{array}
$$

Since $\frac{1}{2}$ is also a term in the sequence, we know $a_{1}=-\frac{1}{4}$. Therefore, $a_{n}=\frac{3}{4} n-1$.