Answer
The sum of the first $n$ terms of the arithmetic sequence is: $ 2n^2 -3n$.
Work Step by Step
The sum of the first $n$ terms of an arithmetic sequence is given by:
$S_{n}= \dfrac{n}{2}\left(a_{1}+a_{n}\right)$
The terms of the sum are the first n terms of the arithmetic sequence, starting with $a_{1}=-1$. The constant difference between the terms is $d=7-3=4$ or, $d=3-(-1)=4$
Now, $S_{n}= \dfrac{n}{2}\left(a_{1}+a_{n}\right) =\dfrac{n}{2}[-1+(4n-5)]\\=\dfrac{n}{2}(4n-6)\\\dfrac{n}{2} \times (2) (2n-3) $
Therefore, the sum of the first $n$ terms of the arithmetic sequence is: $ 2n^2 -3n$.
