Let $a_{1}, a_{2}, a_{3}, \dots be an A.P. If a_{7} = 3$ the product $a_{1} a_{4}$ is minimum and the sum of its first $n$ terms is zero, then $n! - 4 a_{n (n + 2)}$ is equal to :

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$\frac{381}{4}$

$\frac{33}{4}$

answer is D.

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Detailed Solution

$\begin{array}{l} a_{7} = 3, a_{1} + 6 d = 3 \\ a_{1} \cdot (a_{1} + 3 d) = min imum \\ a_{1}^{2} + a_{1} (\frac{3 - a_{1}}{2}) is min \\ a_{1}^{2} + \frac{3 a_{1} - a_{1}^{2}}{2} = \frac{a_{1}^{2} + 3 a_{1}}{2} is min \\ \Rightarrow a_{1} = \frac{- 3}{2} \\ d = \frac{3 - a_{1}}{6} = \frac{3 + \frac{3}{2}}{6} = \frac{9}{2} \times \frac{1}{6} = \frac{3}{4} \end{array}$
Sum of n terms $= 0 \Rightarrow\Rightarrow \frac{n}{2} (2 \cdot (\frac{- 3}{2}) + (n - 1) \cdot \frac{3}{4}) = 0$
$\begin{array}{l} \Rightarrow \frac{n}{2} (\frac{- 12 + 3 n - 3}{4}) = 0 (n > 0) \\ \Rightarrow n = 5 \\ G \cdot E = 5! - 4 a_{5.7} = 120 - 4 \cdot a_{35} = 120 - 4 \cdot (\frac{- 3}{2} + 34 \cdot \frac{3}{4}) = 120 - 96 = 24 \end{array}$