Can the sum of the first n terms of the AP 8, 13, 18, 23, ... ever be exactly 100?

$a=8,d=5$, so $S_n=\frac{n}{2}$$[16+5(n-1)]=\frac{5n^2+11n}{2}$. Setting this equal to 100 gives $5n^2+11n-200=0$, whose discriminant 121 + 4000 = 4121 is not a perfect square $(64^2=4096$ and $65^2=4225$). Numerically $S_5=90$ and $S_6=123$, so the running total jumps straight over 100. A sum of exactly 100 is therefore impossible, because $n$ must be a whole number.