For the AP 27, 24, 21, 18, ... find the value of n (with n > 0) for which the sum of the first n terms is zero.

$a=27,d=-3$, so $S_n=\frac{n}{2}$$[54-3(n-1)]=\frac{n}{2}$$(57-3n)$. This vanishes when $n=0$, which is rejected as trivial, or when $57-3n=0$, that is $n=19$. The tempting answer 10 is the term that is zero $(a_{10}=27-27=0)$, not the point where the running total is zero — the positive terms are only cancelled out much later. $S_{18}=27$ and $S_{20}=-30$, so neither of those is zero either.