find the sum of the first 30 positive integer divisible by 6
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2 thoughts on “find the sum of the first 30 positive integer divisible by 6<br />​”

1. Step-by-step explanation:

   6,12,18,24,30

   a=6

   d=6

   n=30

   Sn=?

   Sn=n/2(2a+(n-1)d

   30/2(2×6+(30-1)6

   15(12+(29)6

   15(12+174)

   15(186)

   2790

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2. Step-by-step explanation:

   Given:–

   First 30 positive integers divisible by 6

   To find:–

   Find the sum of the first 30 positive integers which are divisible by 6 ?

   Solution:–

   The list of positive integers

   = 1,2,3,4,…

   The list of positive integers which are divisible by 6

   = 6,12,18,24,30,…

   The list of first 30 positive integers which are divisible by 6

   = 6,12,18,…(30 terms)

   First term = a = 6

   Common difference = 12-6=6

   Since the common difference is same throughout the series

   6,12,18,…are in the AP

   Now we have to find the sum of first 30 positive integers which are divisible by 6

   = 6+12+18+…+(30 terms)

   We know that

   The sum of first n terms in an AP

   = Sn = (n/2)[2a+(n-1)d]

   We have

   a = 6

   d = 6

   n = 30

   Now,

   S 30 = (30/2)[2(6)+(30-1)(6)]

   => S 30 = (15)[12+29(6)]

   => S 30 = (15)(12+174)

   => S 30 = 15(186)

   => S 30 = 2790

   Answer:–

   The sum of first 30 positive integers which are divisible by 6 is 2790

   Used formulae:–

   The sum of first n terms in an AP

   Sn = (n/2)[2a+(n-1)d]

   • n = number of terms
   • a = First term
   • d = Common difference

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