2. Indicate all values of $n$ for which the sum of $n$ consecutive terms of the arithmetic progression $25,22,19, \ldots$, starting from the first, is not less than 66.
Solution: For the given arithmetic progression, we have: $a_{1}=25, d=-3$. The inequality $S_{n} \geq 66$ must be satisfied, i.e., $\frac{2 a_{1}+d(n-1)}{2} \cdot n \geq 66$. Substituting $a_{1}$ and $d$ into the last inequality, we get

$$
\frac{50-3(n-1)}{2} \cdot n \geq 66 \Leftrightarrow 53 n-3 n^{2} \geq 132 \Leftrightarrow 3 n^{2}-53 n+132 \leq 0 \Leftrightarrow(n-3)\left(n-\frac{44}{3}\right) \leq 0
$$

Since $n$ is a natural number, then $n \in\left[3,14 \frac{2}{3}\right] \cap N$, i.e., $n=3,4,5, \ldots, 14$.

Answer: $3,4,5, \ldots, 14$.