Solution:
the first 40 positive integers divisible by 6 are 6,12,18,....... upto 40 terms
the given series is in arthimetic progression with first term a=6 and common difference d=6
sum of n terms of an A.p is
n/2×{2a+(n-1)d}
→required sum = 40/2×{2(6)+(39)6}
=20{12+234}
=20×246
=4920
