Question

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.

Answer 1

The sequence formed by the given number is
103, 107, 111, 115, ….999.
This is an AP is which a = 103 and d = (107-103) = 4.
Let the total number of these terms be n. Then,
`T_(n) = 999 rArr a +(n-1)d = 999`
`rArr 103 + (n-1) xx 4 = 999`
`rArr (n-1) xx 4 = 896 rArr (n-1) = 224 rArr n = 225 `
`therefore "middle term" = ((n+1)/(2))"th term" = ((225 +1)/(2))`the term = 113th term.
`T_(113) = (a+ 112d) = (103 + 112 xx 4) = 551`
`rArr T_(112) = (551-4) = 547.`
So, we have to find `S_(112) " and " (S_(225) -S_(113))`
Using the formula `S_(m) = (m)/(2)(a+l)` for each sum, we get
`S_(112) = (112)/(2) (103 + 547) = (112 xx 325) = 36400.`
`(S_(225) -S_(113)) = (225)/(2) (103 + 999) - (113)/(2) (103 + 551)`
` = (225 xx 551) - (113 xx 327)`
` = 123975 - 36951 = 87024.`
Sum of all numbers on LHS of the middle term is 36400.
Sum of all numbers on RHS of the middle term is 87024.