In an elevator test, the first cycle has 240 kg load and 15 kg is reduced in each next cycle. In which cycle will the load be 75 kg?

Correct answer: 12th

The situation forms an arithmetic progression because the load changes by the same amount, 15 kg, after every cycle. The first load is a = 240 and the common difference is d = −15. Let the load in cycle n be 75 kg. Applying aₙ = a + (n − 1)d gives 75 = 240 − 15(n − 1). Thus 15(n − 1) = 240 − 75 = 165. Dividing by 15 gives n − 1 = 11, so n = 12. Therefore option C, the 12th cycle, is correct. A direct check confirms the result: cycle 10 has 240 − 9(15) = 105 kg, cycle 11 has 90 kg, cycle 12 has 75 kg, and cycle 13 has 60 kg. The negative difference must be retained because the load decreases rather than increases.