4.

S=(a2+a4)/2*n
3=(a2+a4)/2*3
a2+a4=2
a3=3-2=1
a1=-5
a2=-2
a3=1
a4=4
(d=3)

right.. but i found 2 solutions:

d = 3 a_1 = -5

d=3,5 a_1 = 8

since -2d ²-d+21 = 0 has 2 valid solutions

process: 

arithmetic series are of the form: a_n = a_1 + (n-1) *d

our both given equations translate to:

a_1 +d + a_1+ 2d + a_1 +3d  = 3 => 3 a_1 + 6d =3  

and

a_1 ( a_1 +3d) = -20  (**)

i solved those equations for d via:

 (**) gives a_1 = 1-2d

plugged into gives:

 -2d ²-d+21 = 0

this quadratic equation has 2 solutions d =3 and d=-3,5

those plugged back into can retrive a_1=-5 and a_1 = 8.. done 

S=(a₁+a_n)/2*n
a₂+a₃+a₄=3
3=(a₂+a₄)/2*3 {summary of a.s.}
a₂+a₄=2 - > a₃=3-2=1