Question

1 . find the sum of the first 40 positive integers divisible by 6

Answer 1

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

I hope

Answer 2

hope it's help u...........