Question

find the sum of all the three digit number each of which leaves the remainder3 when divided by 5

Answer 1

it will be an A.P

103,108,..........,998,

here
a=103, d=5, tn=998,

then

998=[103+(n-1)5],

998-103=(n-1)5,

895=(n-1)5,

895/5 =(n-1),

then

n-1=179,

then

n=179+1=180,

therefore

sum=180/2 [2×103+(180-1)5],

=90[206+179×5],

=90[206+895],

=90×1101,

sum=99090