Discussion
What is the sum of all positive integers from 1 to 100, inclusive, that are divisible by 3?
Answer: A
The numbers are 3, 6, 9, ..., 99. This is an AP.
First term a = 3, last term l = 99.
Number of terms n = \(99/3 = 33\).
Sum = \(\frac{n}{2}(a+l) = \frac{33}{2}(3+99) = \frac{33}{2}(102) = 33 \times 51 = 1683\).
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