Question

The sum of all the three digit numbers which when divided by 8 give a remainder of 5 is___

Answer 1

For this condition,
Series is 109,117,125,..........997
a= 109 ,d=8 , n=112
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$=\frac{112}{2}[2\times 109+(112-1\left)8\right]$
=61936.