Two APs both begin with the term 25. The first has common difference 3 and the second has common difference 5.
After how many terms do their sums differ by 210?

The difference of the sums is $\frac{n}{2}$$[50+5(n-1)]-\frac{n}{2}$$[50+3(n-1)]=\frac{n}{2}$$[2(n-1)]=n(n-1)$; notice the equal first terms cancel completely. Setting $n(n-1)=210$ gives $n^2-n-210=0$, so $(n-15)(n+14)=0$ and $n=15$ or $n=-14$. The root -14 is rejected because the number of terms cannot be negative, so $n=15$. At $n=14$ the gap is only 182.