Solution:

Let $a_1, a_2, \ldots, a_{20}$ be the arithmetic sequence, and let $d$ be its common difference. Then $a_1 + a_2 + \cdots + a_{10} = 160$ and $a_1 + a_2 + \cdots + a_{10} + a_{11} + a_{12} + \cdots + a_{20} = 160 + 340 = 500$.

Recalling the formula for the sum of an arithmetic series involving the first term $a_1$ and common difference $d$, the first equation yields $5(2a_1 + 9d) = 160$ or $2a_1 + 9d = 32$, while the second equation yields $10(2a_1 + 19d) = 500$ or $2a_1 + 19d = 50$.

Thus, we get a system of linear equations:
$$\left\{\begin{aligned} 2a_1 + 9d & = 32 \\ 2a_1 + 19d & = 50 \end{aligned}\right.$$
Solving the system gives the value of $a_1$.