Correct Answer - B
Let p and `(p+1)` be removed numbers from `1,2,3,"……"n`, then Sum of the remaining numbers
`=(n(n+1))/(2)-(2p+1)`
From given condition, `(105)/(4)=((n(n1))/(2)-(2p+1))/((n-2))`
`implies 2n^(2)-103n-8p+206=0`
Since, n and p are integers, so n must be even.
Let `n=2r`, we get
`p=(4r^(2)+103(1-r))/(4)`
Since, p is an integer, then `(1-r)` must be divisible by 4.
Let `r=1+4t`, we get
`n=2+8t` and `p=16t^(2)-95t+1`
Now, `1le pltn`
`implies 1le16t^(2)-95t+1lt8t+2`
`implies t=6 " " implies n=50` and `p=7`.
Sum of all numbers `=(50(50+1))/(2)=1275`.
