An AP has 24 terms. The sum of all 24 terms is 1860 and the sum of the first 12 terms is 570.
Find the sum of the last 12 terms.

The last 12 terms are exactly what remains when the first 12 are removed from all 24, so the answer is $S_{24}-S_{12}=1860-570=1290$. The value 2430 adds the two sums instead of subtracting, but $S_{24}$ already contains $S_{12}$. 620 halves $S_{24}$, as if the two halves of an AP had equal sums, which is false whenever d is not 0. 930 comes from an arithmetic slip in the subtraction.