For an AP, the sum of the first 5 terms is 65 and the sum of the first 10 terms is 205.
Find the sum of its first 15 terms.

$S_5=\frac{5}{2}$$(2a+4d)=65$ gives $a+2d=13$, and $S_{10}=5(2a+9d)=205$ gives $2a+9d=41$. Substituting $a=13-2d$ gives $5d=15$, so $d=3$ and $a=7$. Then $S_{15}=\frac{15}{2}$[14 + 14(3)] = $\frac{15}{2}$(56) = 420. The trap 270 adds the two given sums, but they overlap — $S_{10}$ already contains $S_5$. 410 doubles $S_{10}$, and 430 fails the check.