11. (20 points) Given an arithmetic sequence where the first term is less than 0, the 100th term is not less than 74, the 200th term is less than 200, and the number of terms in the interval $\left(\frac{1}{2}, 5\right)$ is 2 less than the number of terms in the interval $\left[20, \frac{49}{2}\right]$. Find the general term formula of the arithmetic sequence.
$$
\begin{array}{l}
\text{Let the first term of the arithmetic sequence be } a\frac{2(l+1)}{9}>\frac{199}{250.74} \\
\Rightarrow \frac{891}{148}-1>l>\frac{1791}{501.48}-1 \\
\Rightarrow 5.03>l>2.57 .
\end{array}
$$

Combining with $3 \mid(l+1)$, we get $l=5$.
Thus, $d=\frac{3}{4}$.
Substituting $l=5$ into equation (7) yields
$$
\begin{array}{l}
\left\{\begin{array}{l}
-50.74<a<0, \\
\frac{74-a}{99} \leqslant \frac{3}{4}<\frac{200-a}{199}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
-50.74<a<0, \\
74-\frac{3 \times 99}{4} \leqslant a<200-\frac{3 \times 199}{4}
\end{array}\right. \\
\Rightarrow-0.25 \leqslant a<0 .
\end{array}
$$

From equation (6), we get
$$
k+4=\frac{5-a}{d}=\frac{4}{3}(5-a) \in\left(\frac{20}{3}, 7\right] .
$$

Thus, $k+4=\frac{5-a}{d}=7 \Rightarrow k=3, a=-\frac{1}{4}$.
Therefore, $a_{n}=-\frac{1}{4}+\frac{3}{4}(n-1)$
$$
=\frac{3}{4} n-1\left(n \in \mathbf{Z}_{+}\right) .
$$
