Question

Let, $S_n$ denote the sum of first $n$ terms of an arithmetic progression whose first term is $-4$ and common difference is $1$. If $V_n=2S_{n+2}-2S_{n+1}+S_n(n\in N)$, then the minimum value of $V_n$ is

• A. $-1$
• B. $9$
• C. $-9$
• D. $-\frac{73}{8}$

Answer 1

The correct option is C $-9$

Given : $V_n=2S_{n+2}-2S_{n+1}+S_n$
$\Rightarrow V_n=2T_{n+2}+S_n\cdots \left(1\right)(\because T_{n+2}=S_{n+2}-S_{n+1})$

The given A.P. has $a=-4,d=1$, so
$T_n=-4+(n-1)=n-5\Rightarrow T_{n+2}=n-3S_n=\frac{n}{2}[-8+(n-1\left)\right]\Rightarrow S_n=\frac{n(n-9)}{2}$
Putting this is equation $\left(1\right)$, we get
$\Rightarrow V_n=2(n-3)+\frac{n(n-9)}{2}\Rightarrow V_n=\frac{n^2-5n-12}{2}$
Minimum of $V_n$ occur at $n=\frac{5}{2}$ but $n\in N$, so checking at near point, i.e. $n=2\text{and}n=3$
At $n=2$, $V_n=-9$
At $n=3$, $V_n=-9$

Hence, the minimum value of $V_n$ is $-9$