1.

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

Answer»

Solution:

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

The answer is 4920


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