2. From the arithmetic sequence $2,5,8,11, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$
2.8 .

First, take $2,5,8,11,20,41,110,1640$, it is easy to see that the sum of their reciprocals is 1, i.e., $k=8$ satisfies the requirement.
Second, suppose we take $x_{1}, x_{2}, \cdots, x_{k}$ from the sequence, such that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1$.
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$

Since $x_{n} \equiv 2(\bmod 3)$, taking the modulus 3 of both sides of equation (1) gives
$k \cdot 2^{k-1} \equiv 2^{k}(\bmod 3)$, i.e., $k \equiv 2(\bmod 3)$.
When $k=2$, $\frac{1}{x_{1}}+\frac{1}{x_{2}} \leqslant \frac{1}{2}+\frac{1}{5}<1$;
When $k=5$,
$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{5}} \leqslant \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}<1$.
Therefore, $k \geqslant 8$. Hence, the minimum value of $k$ is 8.