Answer
$${a_1} = 10,\,\,\,\,d = \frac{2}{3}$$
Work Step by Step
$$\eqalign{
& {S_{31}} = 620,\,\,\,{a_{31}} = 30 \cr
& {\text{Use the first formula for }}{S_n} \cr
& {S_n} = \frac{n}{2}\left( {{a_1} + {a_n}} \right) \cr
& {\text{Let }}n = 31 \cr
& {S_{31}} = \frac{{31}}{2}\left( {{a_1} + {a_{31}}} \right) \cr
& 620 = \frac{{31}}{2}\left( {{a_1} + 30} \right) \cr
& {\text{Solve for }}{a_1} \cr
& 40 = {a_1} + 30 \cr
& {a_1} = 10 \cr
& \cr
& {\text{Now find }}d,{\text{ use the formula for the }}n{\text{th term}} \cr
& {a_n} = {a_1} + \left( {n - 1} \right)d \cr
& {a_{31}} = {a_1} + \left( {31 - 1} \right)d \cr
& 30 = 10 + 30d \cr
& 20 = 30d \cr
& d = \frac{2}{3} \cr
& \cr
& {\text{Then,}} \cr
& {a_1} = 10,\,\,\,\,d = \frac{2}{3} \cr} $$
