find the sum of first 40 positive integers divisible by 6
AP : 6, 12, 18,...,(6 x 40).
a = 6
d = 6
an = 6 x 40 = 240
Sn = n/2 [2a + (n-1) d]
=40/2 [ 2 x 6 + (40 - 1) 6 ]
=20 [12 + 39 x 6]
=20 [12 + 234]
=20 x 246
=4920
Therefore, the sum of the first 40 positive integers divisible by 6 = 4920.
and I assure you this is correct...!...=D...