In the AP (12,17,22,\ldots), find the sum from the (21)st term to the (40)th term.

Correct answer: (3190)

In an arithmetic progression, each term changes by the same common difference. Here the first term is 12 and the common difference is 5, so the term at position n is found from the rule \(a_n=12+(n-1)5\). The requested terms run from the 21st through the 40th, giving 20 terms in all. Their first term is \(a_{21}=112\), and their last term is \(a_{40}=207\). The sum of an AP can be found by multiplying the number of terms by the average of the first and last terms.

Thus, the required sum is \(20\times\frac{112+207}{2}=10\times319=3190\). Equivalently, one may calculate \(S_{40}-S_{20}\), because subtracting the first 20 terms leaves terms 21 to 40. Therefore option A is correct. A common error is to subtract \(S_{21}\), which would remove the 21st term too and give the wrong range.