Question

find the sum of first 40 positive integers which is divisible by 6?​

Answer 1

Hi friend,

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 this will help u ;)

Answer 2

a=6

d=6

n=40

Sn=n/2[2a+(n-1)]d

S40=40/2[2×6+(40-1)]6

S40=20(12+39)×6

S40=20(12+234)

S40=20×246

S40=4920