Question

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

Answer:

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

This is an example of A.P. (Arithmetic progression).

Sum of n terms of an A.P. is given by,

Sn=n/2×[2a+(n-1)d]

where,

a=first term =6

d=common difference=6

Thus,

Sn = 40/2×{2(6)+(39)6}

=20{12+234}

=20×246

=4920

Hence The Answer is 4920