7. From the arithmetic sequence $2,5,8, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ ـ.
7.8.

First, let's take $x_{1}, x_{2}, \cdots, x_{k}$ from the known 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
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$

It is easy to see that for any $n$, $x_{n}=2(\bmod 3)$.
Thus, taking modulo 3 on both sides of equation (1) gives
$$
2^{k-1} k \equiv 2^{k}(\bmod 3) \Rightarrow k \equiv 2(\bmod 3) \text {. }
$$

When $k=2$,
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}} \leqslant \frac{1}{2}+\frac{1}{5}<1 \text {; }
$$

When $k=5$,
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{5}} \leqslant \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}<1 \text {; }
$$

When $k=8$, taking $2, 5, 8, 11, 20, 41, 110, 1640$ satisfies the condition.