Question

find the middle term of the sequence formed bya all three digit number which leave a remainder 3 and when divided by 4.also find sum of all numbers on both sides of the middle term seperately..

Answer 1

The English grammar and sentence of this question is not clear. So it takes a long time to understand the question.

Clue: All Three digit numbers divided by 4 gives 3 as REMAINDER 

So the numbers are somewhere between 100 till 999 (all are three digits) 

100 is exactly divisible by 4, 101 leaves remainder 1, 102 leaves remainder 2, 103 leaves remainder 3, 104 is exactly divisible by 4, 105 leaves remainder 1, 106 leaves remainder 2, 107 leaves remainder 3, 108 is exactly divisible by 4 and so on......... 

From the other end 999 leaves remainder 3 when divided by 4. So that is the last term 

So the sequence starts from 103,107,111..........999 

999 is the last term 

To find out total number of terms in the progression use formula 999= a + (n-1) d
So 999=103+(n-1)X4 

Therefore n=225 

There are 112 numbers before and 112 numbers and 113th number is = 103+(113-1)X4=551 


Sum of number before 551 

There are 112 numbers starting with 103 
Answer is 36400