Question

find sum of all natural numbers 1 to 125 which are divisible by 3​

Answer 1

Answer:

2583

Step-by-step explanation:

3,6,9......123

123 = a + (n-1)d

123 = 3 + (n-1)3

120 = 3n - 3

n = 123/3 = 41

sum = 41/2(6 + 120)

      41/2 * 126 =  2583