Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 3, when divided by 4. Also find the sum of all numbers on both sides of the middle terms separately

Explanation

The sequence of three-digit numbers which leave a remainder 3, when divided by 4 is `103,107,111,115,119,123,... 999`
Here, first term a=103, d=107-103=4
Let `T_(n)=999`
`rArr a+(n-1)d=999`
`rArr 103+(n-1)4=999 rArr (n-1)4=896`
`:. n-1=224 rArr n=225`
Therefore , `(225+1)/(2)`=113th term is the middle term.
`:. ` Middle term `=T_(113)=a+112d=103+112(4)=551`
`:.` First term after the middle term `=T_(114)=a+113d=103+113(4)=555`
`:.` Sum of first 112 term `=(112)/(2)(2xx103+111xx4)=36,400`
and ` " " `sum of last 112 term `=(112)/(2)(2xx555+111xx4)=87,024`