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Question

The sum of first 40 positive integers divisible by 6 is
(a) 2460   (b) 3640   (c) 4920    (d) 4860
 


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


 

Mathematics
Secondary School Mathematics X
Standard X

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