Question

find the middle term sequence formed by all 3 digit numbers which leave remainder 4 also find the sum of all numbers on both sides of middle trrm separately

Answer 1

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, an = 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.

middle term = (n+12)th term = 113th term = a + 112d = 103 + 112×4 = 551

We know that, sum of first n terms of an AP is,Sn = n2[2a+(n−1)d]

Now, Sum = 112/2[2×103 + 111×4] = 36400

Sum of all terms before  middle term  = 36400

sum of all numbers=  225/2[2×103+224×4] = 123975

Now, sum of terms after  middle term = S225 − (S112+551) = 123975−(36400+551) = 87024