In an arithmetic sequence each term is the sum of the previous term and a fixed ratio. For this case we have the 11th term and the 15th term, so there are 4 steps between the two terms, which means that the 11th term plus 4 times the ratio should be equal to the 15th one.
[tex]\begin{gathered} a15=a11+4\cdot r \\ -1=31+4\cdot r \\ 4\cdot r=-1-31 \\ 4\cdot r=-32 \\ r=\frac{-32}{4} \\ r=-8 \end{gathered}[/tex]The ratio is -8. We can use the same idea to determine the 28th term, there are 13 steps between the 15th term and the 28th one. So we have:
[tex]\begin{gathered} a28=a15+13\cdot r \\ a28=-1+13\cdot(-8) \\ a28=-1-104=-105 \end{gathered}[/tex]The answer is -105.
