Question

find the sum of the first 40 positive integers divisible by 6

Answer 1

AP will be 6 , 12 , 18 ......
n: 40
An = a+ (n -1)d
6+ 39*6
6+234
An= 240
now , Sn : n/2(a+l)
40/2(6+240)
20*246
Sn = 4920

Answer 2

40 positive integers divisible by 6 are 6, 12, 18, ........ 40 terms .
here we have to use the concept of AP.
Let first term (a ) = 6 ,
common difference (d ) = 6
and no of terms (n ) = 40
$T_n=a+(n-1)d\\\\=6+(40-1) \times 6=240$

now, use formula,
$S_n=\frac{n}{2}[first term + nth term]\\\\=\frac{40}{2}[6+240]\\\\=20\times\:246=4920$