2. Four positive integers form a monotonically increasing sequence, where the first three terms form an arithmetic sequence, and the last three terms form a geometric sequence. If the difference between the last term and the first term is 30, then the sum of these four positive integers is $\qquad$ .
2. 129.

Let the terms of the sequence be $a, a+d, a+2d, a+30\left(a, d \in \mathbf{N}^{*}\right)$, then
$$
(a+30)(a+d)=(a+2d)^{2} \Rightarrow 3a(10-d)=2d(2d-15).
$$

Thus,
$$
(10-d)(2d-15)>0 \Rightarrow \frac{15}{2}<d<10.
$$

Therefore, $d=8, a=\frac{8}{3}$ (discard) or $d=9, a=18$.
Hence, the sequence is $18, 27, 36, 48$. The required sum is 129.