Question

find the sum of the first 40 integers divisible by 6 in arithmetic progression

Answer 1

6,12,18.....
n=40
a=6,d=6
n/2 [2a + (n-1)d]
40/2[2(6) +(40-1)6]
20(12+39×6)
20(12+234)
20(246)
4920 is answer