help with math homework?

Let,
first-term = a,
common-difference = d,

S12 = (12/2)[2a + 11d] = 282,
2a +11d = 282*2/12 = 47,

2a = 47-11d ..................................................................... [1]
Also,
S19 = (19/2)[2a +18d] = 646,
2a +18d = 646*2/19 = 68

2a = 68 -18d ........................................................... [2]

From [1] and [2],
47-11d = 68-18d, ===> -11d +18d = 68 -47,
7d = 21,

d = 21/7 = 3 >================================< ANSWER
and
From [1], 2a = 47-11*3 = 47 -33 = 14,

a = 7 >=======================================< ANSWER

an=a1+(n-1)d and Sn=(n(a1+an)) / 2

282=(12(a1+a12)) / 2

646=(19(a1+a19)) /2

Solved, you will get two equations, 47=a1+a12
68=a1+a19

Substitute a12=a1+11d into 47=a1+a12
Substitute a19=a1+18d into 68=a1+a19

Now you have two simultaneous equations. 47=2a1+11d
-68=2a1+18d

Solve it you should get d=3 and then substitute back into the equation and find a1=7
Therefore a1=7 and d=3. Hope it helps and good luck.