In an AP (a_{20}=178) and (a_{52}=626). What is (a_{36})?

Correct answer: (402)

The direct answer is option C, 402. In an arithmetic progression, equal increases or decreases occur between consecutive terms. The terms a20 and a52 have indices 20 and 52, and 36 is exactly halfway between them because 36−20=16 and 52−36=16. Therefore, the middle term is the average of the two end terms: a36=(a20+a52)/2=(178+626)/2=804/2=402. Option A, 382, is too small and does not equal this average. Option B, 392, is also below 402. Option C, 402, is correct because the required index is exactly midway between 20 and 52. Option D, 412, is too large. A useful exam cue is: in an AP, if the indices are equally spaced, the corresponding terms are also equally spaced, so the middle term is the average of the outer terms.