Question

find the sum of 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

Answer 2

40 positive integers divisible by 6 are

6,12,18.......,240


thus

s = 6+12+18+....+240

s = 6(1+2+3....+40)

sum of n natural number = n(n+1)/2

s = 6*(40)(41)/2

s = 4920