+118608 sum of all multiples of 6 between 6 and 999
This is an arithemetic progression.
The first term is 6*2=12 and the last term is 6*166=996
So that is 166-2+1=165 term
T1=6
n=165
T165= 996
d=6
\(Sn=\frac{n}{2}(T_1+Last)\\ S_{165}=\frac{996}{2}(6+996)\\ S_{165}=\frac{996}{2}(1002)\\ \)
sum = 996/2*1002 = 498996
