Q. What is the number of terms in the series 117,120,123,126, ..., 333?
Answer: 73
Notes: The given series 117, 120, 123, 126, …, 333 is an arithmetic progression where the first term (a) is 117 and the common difference (d) is 3. The last term of the series is 333. Using the formula for the last term of an arithmetic progression, l = a + (n − 1)d, we substitute the values: 333 = 117 + (n − 1) × 3. Subtracting 117 from both sides gives 216 = 3(n − 1). Dividing both sides by 3 gives n − 1 = 72, so n = 73. Therefore, the number of terms in the series is 73 (Option B).