Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots$ be an arithmetic sequence. Find $a_{10}$ and $S_{20}$
\[
a_{1}=17, d=3
\]
\[
a_{10}=
\]
(Simplify your answer.)

Step 1 ：Given an arithmetic sequence where the first term \(a_1 = 17\) and the common difference \(d = 3\).

Step 2 ：We are asked to find the 10th term of the sequence, \(a_{10}\).

Step 3 ：In an arithmetic sequence, the nth term can be found using the formula \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

Step 4 ：Substitute \(a_1 = 17\), \(d = 3\), and \(n = 10\) into the formula to find \(a_{10}\).

Step 5 ：\(a_{10} = 17 + (10-1)\times3 = 44\).

Step 6 ：We are also asked to find the sum of the first 20 terms of the sequence, \(S_{20}\).

Step 7 ：The sum of the first n terms of an arithmetic sequence can be found using the formula \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(a_1\) is the first term, \(a_n\) is the nth term, and \(n\) is the number of terms.

Step 8 ：Substitute \(a_1 = 17\), \(a_n = 44\), and \(n = 20\) into the formula to find \(S_{20}\).

Step 9 ：\(S_{20} = \frac{20}{2}(17 + 44) = 610\).

Step 10 ：Final Answer: The 10th term of the arithmetic sequence is \(\boxed{44}\) and the sum of the first 20 terms of the sequence is \(\boxed{610}\).