In an AP, a₁₃ = 65 and a₂₁ = 113. What is a₆?

Correct answer: 23

For an arithmetic progression, the change between two terms equals the difference in their indices multiplied by the common difference. From the 13th to the 21st term there are 21 − 13 = 8 equal intervals. Hence a₂₁ − a₁₃ = 113 − 65 = 48 = 8d, giving d = 6. The sixth term is seven intervals before the thirteenth term because 13 − 6 = 7. Moving backward therefore gives a₆ = a₁₃ − 7d = 65 − 7(6) = 65 − 42 = 23. Thus option C is correct. This can also be checked by reconstructing the terms: a₁₃ = a₁ + 12d, so a₁ = 65 − 72 = −7 and a₆ = −7 + 5(6) = 23. Adding instead of subtracting when moving backward, or using the wrong interval count, produces the distractors.