JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 7)
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 :
24
$\frac{381}{4}$
9
$\frac{33}{4}$
Explanation
$a_{7}=3 \Rightarrow a+6 d=3 \Rightarrow a=3-6 d$
$$ \begin{aligned} & a_{1} \cdot a_{4}=a(a+3 d) \\\\ & \Rightarrow(3-6 d)(3-6 d+3 d) \\\\ & \Rightarrow 3(1-2 d) 3(1-d) \\\\ & \Rightarrow 9\left(2 d^{2}-3 d+1\right) \end{aligned} $$
Let $f(d)=2 d^{2}-3 d+1$
$f^{\prime}(d)=4 d-3 \Rightarrow d=\frac{3}{4}$
$\therefore a=3-6 \cdot \frac{3}{4}=3-\frac{9}{2}=-\frac{3}{2}$
$S_{n}=0$
$\frac{n}{2}(29+(n-1) d)=0$
$\Rightarrow 2 \cdot\left(-\frac{3}{2}\right)+(n-1)\left(\frac{3}{4}\right)=0$
$\Rightarrow \quad 3=\frac{3}{4}(n-1)$
$\Rightarrow n=5$
Now, $n !-4 \cdot a_{n(n+2)}$
$$ \begin{aligned} & =5 !-4 \cdot a_{35} \\\\ & =120-4\left(-\frac{3}{2}+34 \cdot \frac{3}{4}\right) \\\\ & =120-(-6+102) \\\\ & =120-(96) \\\\ & =24 \end{aligned} $$
$$ \begin{aligned} & a_{1} \cdot a_{4}=a(a+3 d) \\\\ & \Rightarrow(3-6 d)(3-6 d+3 d) \\\\ & \Rightarrow 3(1-2 d) 3(1-d) \\\\ & \Rightarrow 9\left(2 d^{2}-3 d+1\right) \end{aligned} $$
Let $f(d)=2 d^{2}-3 d+1$
$f^{\prime}(d)=4 d-3 \Rightarrow d=\frac{3}{4}$
$\therefore a=3-6 \cdot \frac{3}{4}=3-\frac{9}{2}=-\frac{3}{2}$
$S_{n}=0$
$\frac{n}{2}(29+(n-1) d)=0$
$\Rightarrow 2 \cdot\left(-\frac{3}{2}\right)+(n-1)\left(\frac{3}{4}\right)=0$
$\Rightarrow \quad 3=\frac{3}{4}(n-1)$
$\Rightarrow n=5$
Now, $n !-4 \cdot a_{n(n+2)}$
$$ \begin{aligned} & =5 !-4 \cdot a_{35} \\\\ & =120-4\left(-\frac{3}{2}+34 \cdot \frac{3}{4}\right) \\\\ & =120-(-6+102) \\\\ & =120-(96) \\\\ & =24 \end{aligned} $$
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