SOLUTION:
We are told that;
[tex]\begin{gathered} a+a+d=15 \\ a+2d+a+3d=43 \end{gathered}[/tex]
Rewriting, we have;
[tex]\begin{gathered} 2a+d=15 \\ 2a+5d=43 \end{gathered}[/tex]
Subtracting the first and second equations, we have;
[tex]\begin{gathered} 4d=28 \\ d=7 \end{gathered}[/tex]
and;
[tex]\begin{gathered} 2a+7=15 \\ 2a=8 \\ a=4 \end{gathered}[/tex]
Thus, the first four terms of the sequence are;
[tex]4,11,18,25[/tex]
