Answer:
Sum:
[tex]{ \bf{S _{n} = \frac{n}{2} (2a + (n - 1)d) }} \\ \\ { \sf{2n(n + 3) = \frac{n}{2}(2a + (n - 1)d) }} \\ \\ { \sf{4 {n} + 12=2a + (n - 1)d }} \\ { \sf{2a = 4n - nd + d + 12}} \\ { \sf{2a = n(4 - d) + 12 + d}} \\ { \sf{a = \frac{1}{2}(n(4 - d) + 12 + d) }}[/tex]
