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Question

Let, Sn denote the sum of first n terms of an arithmetic progression whose first term is 4 and common difference is 1. If Vn=2Sn+22Sn+1+Sn (nN), then the minimum value of Vn is 


Your Answer
A
1
Your Answer
B
9
Correct Answer
C
9
Your Answer
D
738

Solution

The correct option is C 9
Given : Vn=2Sn+22Sn+1+Sn
Vn=2Tn+2+Sn(1)    (Tn+2=Sn+2Sn+1)

The given A.P. has a=4, d=1, so 
Tn=4+(n1)=n5Tn+2=n3Sn=n2[8+(n1)]Sn=n(n9)2
Putting this is equation (1), we get
Vn=2(n3)+n(n9)2Vn=n25n122
Minimum of Vn occur at n=52 but nN, 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

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