Question

Find sum of all 3 digit number which are divisible by 6 and 4 both

Answer 1

first 3 digit number divisible by both 6 & 4 is 108

last 3 digit number divisible by both 6 & 4 is 996

total number of terms is 75

Sn = n/2(2a+(n-1)d)

$= \frac{75}{2} (2 \times 108 + (75 - 1)12) \\ \frac{75}{2} (216 + (74 \times 12)) \\ \frac{75}{2} (216 + 888) \\ \frac{75}{2} \times 1104 \\ 75 \times 552 \\ = 41400$