$$\begin{gathered} \underline{\begin{array}{c}\\ {}_{+}\end{array}\{\begin{array}{l}{a}_1+3r=12\\ -a_1-16r=-64\end{array}} \\ -13r=-52|:(-13) \\ r=4 \\ \{\begin{array}{l}r=4\\ a_1+3r=12\end{array} \\ \{\begin{array}{l}r=4\\ a_1+3\cdot 4=12\end{array} \\ \{\begin{array}{l}r=4\\ a_1+12=12|-12\end{array} \\ \{\begin{array}{l}r=4\\ a_1=0\end{array} \end{gathered}$$

Czyli wzór ogólny tego ciągu arytmetycznego to:

$a_n=0+\left(n-1\right)\cdot 4=4n-4$