7、The arithmetic sequence $a_{1}, a_{2}, \cdots, a_{19}$ has a total of 19 terms. It is known that $a_{1}+a_{7}+a_{14}+a_{18}=120$, then $a_{1}+a_{2}+a_{3} \cdots+a_{19}=$ $\qquad$
[Answer] 570
[Solution] Let the common difference of the arithmetic sequence be $d$, then $\left\{\begin{array}{l}a_{7}=a_{1}+6 d \\ a_{14}=a_{1}+13 d \\ a_{18}=a_{1}+17 d\end{array}\right.$, so
$$
\begin{aligned}
a_{1}+a_{7}+a_{14}+a_{18}= & 4 a_{1}+36 d=120 \Rightarrow a_{1}+9 d=30 \text { . And } \\
a_{1}+a_{2}+a_{3} \cdots+a_{19} & =a_{1}+\left(a_{1}+d\right)+\left(a_{1}+2 d\right) \cdots+\left(a_{1}+18 d\right) \\
& =19 a_{1}+\frac{(1+18) \times 18}{2} d \\
& =19 a_{1}+171 d \\
& =19\left(a_{1}+9 d\right) \\
& =19 \times 30=570
\end{aligned}
$$