The sum of the first n terms of the AP 8, 12, 16, 20, ... equals the sum of the first n terms of the AP 17, 19, 21, 23, ....
Find n, taking n > 0.

First AP: $a=8,d=4$, so $S_n=\frac{n}{2}$$[16+4(n-1)]=2n^2+6n$. Second AP: $a=17,d=2$, so $S_n=\frac{n}{2}$$[34+2(n-1)]=n^2+16n$. Equating gives $2n^2+6n=n^2+16n$, so $n^2-10n=0$ and $n(n-10)=0$. The root $n=0$ is rejected because a sum of no terms is trivial, leaving $n=10$. (Both sums are then 260; at $n=9$ they are 216 and 225, so 9 is too small.)