a) We have that a1 = 2 and a4 = 14, so the first thing that we can conclude is that the common difference, d, is positive, and that a2 and a3 are in between 2 and 14. We know that a1 + 3d = 14. Substituting a1 = 2 into the equation and rearranging, we get that 3d = 12, which can be further simplified to get d = 4.
b) There is a formula which you can use to get the answer quickly, and this formula comes in 2 forms:
Sn = 0.5n(2a1 + (n − 1)d) OR Sn = 0.5n(a + ℓ), where ℓ refers to the last term in the range that we are considering.
Either formula will work, but in this case we will use the first form. Making the appropriate substitutions, we end up with
S14 = 0.5(14) [2(2) + (14-1)(4)]
Upon simplifying, we get S14 = 7 [4 + 52] = 392, which is the final answer that w are looking for.