jmc

Let the first term of the arithmetic sequence be $a$, and let the common difference be $d$. Then the second term is $a+d=17$, fifth term is $a+4d=19$, and the eighth term is $a+7d$. Note that $(a+4d)-(a+d)=3d$, and $(a+7d)-(a+4d)=3d$, so the terms $a+d=17$, $a+4d=19$, and $a+7d$ also form an arithmetic sequence.

If 17 and 19 are consecutive terms in an arithmetic sequence, then the common difference is $19-17=2$, so the next term must be $19+2=\boxed{21}$.