Question

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

Answer 1

The positive integers that are divisible by 6 are
$6, 12, 18, 24 ….$
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
$a = 6$ and $d = 6$
$S_{40 }=?$
$S_n = \frac n2 [2a + (n-1)d]$

$S_{40} = \frac {40}{2} [2(6) + (40-1)6]$
$S_{40} = 20[12 + (39) (6)]$
$S_{40} = 20(12 + 234)$
$S_{40} = 20 × 246$
$S_{40 }= 4920$