1. Given the inequality
$$
3(x-1)<2(x+a)-5
$$

has only three positive integer solutions. Then the range of $a$ is ( ).
(A) $\frac{5}{2} \leqslant a<3$
(B) $\frac{5}{2} \leqslant a \leqslant 3$
(C) $\frac{5}{2}<a \leqslant 3$
(D) $3<a<\frac{7}{2}$
$-1 . C$.
From the given, we have $0<x<2 a-2$.
Combining the problem statement, we get
$$
3<2 a-2 \leqslant 4 \Rightarrow \frac{5}{2}<a \leqslant 3 .
$$