If the sum of the first n terms of an Arithmetic Progression is equal to $n^2 + 3n$

Question

If the sum of the first n terms of an Arithmetic Progression is equal to $n^2 + 3n$ then the first term of the Arithmetic Progression is I tried to solve this question by putting sum of of $n^th$ term of an Arithmetic Progression that is $(n÷2)(2a+(n-1)d)$= $n^2+3n$ but won't end up with an answer.please help me.

Answer 1

Taking $n=1$ we see that $S_1=a_1=4$.

Answer 2

From your formula we get $$\frac{n(2a_1+(n-1)d)}{2}=n^2+3n$$ for $$n\neq 0$$ we get $$2a_1=2n+6-(n-1)d$$ so...?