The correct option is C −9
Given : Vn=2Sn+2−2Sn+1+Sn
⇒Vn=2Tn+2+Sn⋯(1) (∵Tn+2=Sn+2−Sn+1)
The given A.P. has a=−4, d=1, so
Tn=−4+(n−1)=n−5⇒Tn+2=n−3Sn=n2[−8+(n−1)]⇒Sn=n(n−9)2
Putting this is equation (1), we get
⇒Vn=2(n−3)+n(n−9)2⇒Vn=n2−5n−122
Minimum of Vn occur at n=52 but n∈N, so checking at near point, i.e. n=2 and n=3
At n=2, Vn=−9
At n=3, Vn=−9
Hence, the minimum value of Vn is −9