find the sum of first 40 positive integers divisible by 6

 we have to find sum of first 40 positive integers divisible by six.....

therefore :

our ap so formed is : 6,12,18 .......240

where a or the first term = 6 .......

and d or the common difference =12-6 = 6

therefore sum of first 40 positive integers divisible by 6 =n / 2 [ 2a + ( n- 1 ) d ]

                                                                                                   = 40 / 2 [ 2 x 6 + ( 5 ) 6 ]

                                                                                                   = 20 [ 12 + 30 ]

                                                                                                  = 20 x 42

                                                                                                   = 840.

AP : 6+12+18+...+36.

a = 6

d = 6

a_n = 36

a_n = a + (n-1) d

36 = 6 + (n-1) 6

36 - 6 = (n-1) 6

30 = (n-1) 6 

(n-1) = 30/6 

(n-1) = 5

n = 5 + 1

n = 6

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

S₆ = 6/2 [2 x 6 + (6-1) 6]

      = 3 [12 + 5 x 6]

      = 3 [12 + 30]

      = 3 [42]

      = 126

Therefore, the sum of the first 40 positive integers divisible by 6 = 126.

hope this helps...!...