Question

pls give me ans fast

Answer 1

The smallest three digit number = 100 but when divided by 5 it leaves remainder = 0

=> 103 is the smallest three digit number which will give remainder = 3 when divided by 5

The last three digit number is 999 which leaves 4 as remainder when divided by 5.

=> 998 will give remainder 3 when divided by 5.
The series is obviously an A.P.,
(103, 108, 111,...,998)

First term,a = 103;
Common difference,d = 5;
an = 998
So,
an = a + (n - 1)d
998 = 103 + (n - 1) 5
998 - 103/5 = n-1
895/5 = n-1
179 + 1 =n
n = 180

Now, their sum...
= 180/2(103 + 998)
= 90×1101
= 99090

Hope it is correct...

Mark it as brainliest please!!!


Regards...