AP ICET 2016 Paper

The first and the last 3-digit numbers divisible by 4 are 100 and 996. So, the 3-digit numbers that would leave a remainder 1 in each case are 101, 105, .............., 997
Total number of such 3-digits numbers = 225
It forms an A.P with n= 225, a = 101, l = 997
∴ Required sum $=\frac{\mathit{n}}{2}\left(\mathit{a}+1\right)=\frac{225}{2}\left(101+997\right)$
$=123525$