The sum of all natural numbers from 100 to 200, which are divisible by 5, is:
Answer: A
The series is 100, 105, ..., 200. This is an A.P. with \(a=100, l=200, d=5\). Number of terms \(n = \frac{l-a}{d} + 1 = \frac{200-100}{5} + 1 = 20+1=21\). Sum = \(\frac{n}{2}(a+l) = \frac{21}{2}(100+200) = \frac{21}{2}(300) = 21 \times 150 = 3150\).