Q. 124.2( 227 Votes )

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

The smallest positive integer which is divisible by 6 is 6 itself and its 40th multiple will be 6 x 40 = 240

So, we have a = 6, d = 6, n = 40 and 40th term = 240.

Sum of first 40 positive integers divisible by 6 can be calculated as follows:

= 20(12 + 234)

= 20*246

= 4920

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