# find the middle term of the sequence formed by all three digits numbers which leave a reminder 3, when divided by 4. Also find the sum of all numbers on both sides of the middle term separately.

The list of 3 digit number that leaves a remainder of 3 when divided by 4 is :

103  ,  107 , 111 , 115 ,   .... 999

The above list is in AP with first term, a = 103 and common difference, d = 4

Let n be the number of terms in the AP.

Now, a= 999

103 + ( n  -  1 ) 4 =  999

103 + 4n  - 4 =  999

4n  + 99 = 999

4n  =  900

n  =  225

Since, the number of terms is odd, so there will be only one middle term.

• 187

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