find the sum of the first 40 posative integers divisible by 6​

2 thoughts on “find the sum of the first 40 posative integers divisible by 6​”

1. Hello guys you can check my answer , but I am not fully sure about the answer .

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2. Answer:

   The first 40 positive integers that are divisible by 6 are 6,12,18,24…

   a=6 and d=6.

   We need to find S

   40

   S

   n

   =

   2

   n

   [2a+(n−1)d]

   S

   40

   =

   2

   40

   [2(6)+(40−1)6]

   =20[12+(39)6]

   =20(12+234)

   =20×246

   =4920

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