Question

I have the first part but I don't know how to do the second part h e l p

Answer 1

$\bf \begin{array}{cll} \stackrel{ordinal}{position}&\stackrel{term}{value}\\ \cline{1-2} 1&12\\ 2&12+d\\ 3&12+d+d\\ 4&12+d+d+d\\ &12+3d \end{array}\qquad \implies ~\hfill \begin{array}{llll} \stackrel{\textit{4th term}}{12+3d}~~=~~\stackrel{\textit{4th value}}{3}\\\\ 3d=-9\implies d = \cfrac{-9}{3}\\\\ \stackrel{\textit{common difference}}{d = -3} \end{array}$

well, we know the common difference is -3, to go from the 4th term to the 8th term, we need to add "d" 4 times or namely 3+4(-3), likewise to go from the 13th term to the 19th term we have to add "d" 6 times, or namely -24 + 6(-3).

$\bf \stackrel{\textit{8th term}}{3+4(-3)}\implies 3-12\implies -9 \\\\\\ \stackrel{\textit{19th term}}{-24 + 6(-3)}\implies -24-18\implies -42$