Question

While learning I came across a question which was as follows:

Find the sum of the first 40 positive integers divisible by 6.

My doubt is that should I find the sum of the 1st 40 multiples of 6 or the multiples of 6 from starting from 1 to 40 (i.e, 6,12,18,24,30,36).

Answer 1

I have explained your doubt as well as the solution for your question.
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