An AP has 15 terms whose total is 600, and its first term is 12.
Find the sum of its last 5 terms.

First find $d:600=\frac{15}{2}$$[24+14d]$, so $80=24+14d$, giving $14d=56$ and $d=4$. The last 5 terms are terms 11 to 15, so their sum is $S_{15}-S_{10}=600-\frac{10}{2}$$[24+36]=600-300=300$. (Check: $a_{11}=52,a_{15}=68$ and $\frac{5}{2}$$(52+68)=300$.) The value 248 subtracts $S_{11}$, wrongly deleting the 11th term too. 290 averages $a_{10}$ and $a_{15}$ over 5 terms, but $a_{10}$ is not one of the last five. 200 assumes the last third of the terms carries a third of the total, which is false whenever $d$ is not 0.