from the 3rd term to the 8th term, we'd need to use the "common difference" or "d", 5 times, namely add it that many times, to get the 8th term, thus
[tex]\bf \begin{array}{ll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_3&-12\\
a_4&-12+d\\
a_5&-12+d+d\\
a_6&-12+d+d+d\\
a_7&-12+d+d+d+d\\
a_8&-12+d+d+d+d+d\\
&-12+5d\\
&-37
\end{array}\implies -12+5d=-37
\\\\\\
5d=-25\implies d=\cfrac{-25}{5}\implies d=-5[/tex]
we know the 3rd term is -12, so the first one has to be -12+5+5, undoing the -5's, so the first term is -2.
[tex]\bf n^{th}\textit{ term of an arithmetic sequence}
\\\\
a_n=a_1+(n-1)d\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=-2\\
d=-5
\end{cases}
\\\\\\
a_n=-2+(n-1)(-5)[/tex]
