Let the consecutive terms be "q" and "q+1".
as per given:
{(n(n+1)/2) - (2q +1)}/(n-2) =105/4
i.e. 2n^2 - 103n - 8q + 206 =0
-8q+206 will be even...so n must be even...
let n=2r..
now q={4r^2 + 103(1-r)}/4
since q is integer so (1-r) must be divisible by 4....
so let r=1+4t
since n=2r..
n=2+8t
q=16t^2 - 95t + 1
1<=q<n
1<=16t^2 - 95t + 1 < 8t+2
i.e t=6
so n=50 and
q=7
therefore removed nos. are 7 and 8