Let $a_1, a_2, a_3, \ldots$ 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 :

$a_{7}=3 \Rightarrow a+6 d=3 \Rightarrow a=3-6 d$ <br/><br/>$$ \begin{aligned} & a_{1} \cdot a_{4}=a(a+3 d) \\\\ & \Rightarrow(3-6 d)(3-6 d+3 d) \\\\ & \Rightarrow 3(1-2 d) 3(1-d) \\\\ & \Rightarrow 9\left(2 d^{2}-3 d+1\right) \end{aligned} $$ <br/><br/>Let $f(d)=2 d^{2}-3 d+1$ <br/><br/>$f^{\prime}(d)=4 d-3 \Rightarrow d=\frac{3}{4}$ <br/><br/>$\therefore a=3-6 \cdot \frac{3}{4}=3-\frac{9}{2}=-\frac{3}{2}$ <br/><br/>$S_{n}=0$ <br/><br/>$\frac{n}{2}(29+(n-1) d)=0$ <br/><br/>$\Rightarrow 2 \cdot\left(-\frac{3}{2}\right)+(n-1)\left(\frac{3}{4}\right)=0$ <br/><br/>$\Rightarrow \quad 3=\frac{3}{4}(n-1)$ <br/><br/>$\Rightarrow n=5$ <br/><br/>Now, $n !-4 \cdot a_{n(n+2)}$ <br/><br/>$$ \begin{aligned} & =5 !-4 \cdot a_{35} \\\\ & =120-4\left(-\frac{3}{2}+34 \cdot \frac{3}{4}\right) \\\\ & =120-(-6+102) \\\\ & =120-(96) \\\\ & =24 \end{aligned} $$