If an arithmetic progression has (S_6=111) and (S_{13}=494), what is the sum of the (7)th to (13)th terms?

Correct answer: (383)

Direct answer: Option D, 383. An AP has partial sums written as S_n, where S_n means the sum of the first n terms. The first six terms have sum S_6=111, while the first thirteen terms have sum S_13=494. The terms from the seventh through the thirteenth are exactly the terms included in S_13 but not included in S_6. Therefore, subtract the earlier sum: S_13-S_6=494-111=383. Option A, 353, is not obtained from the given subtraction. Option B, 363, is also an incorrect difference. Option C, 373, is another arithmetic distractor. Option D, 383, is correct because it represents precisely terms 7 through 13, inclusive. Memory cue: for a consecutive block beginning after the first m terms, use the larger partial sum minus the earlier partial sum.