For the AP 6, 10, 14, 18, ... find the sum of the terms from the 11th to the 25th (both included).

The wanted sum is $S_{25}-S_{10}$. With $a=6$ and $d=4,S_{25}=\frac{25}{2}$$[12+24(4)]=25(54)=1350$ and $S_{10}=5[12+36]=240$, so the answer is 1350 - 240 = 1110. (Check: $a_{11}=46,a_{25}=102$, 15 terms, $\frac{15}{2}$$(46+102)=1110$.) The value 1350 forgets to remove the first ten terms; 1064 subtracts $S_{11}$, which wrongly deletes the 11th term as well; 1036 counts only 14 terms.