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Q.
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 +6 d =3 $.......................(1)
$ Z = a ( a +3 d ) $
$ =(3-6 d )(3-3 d )$
$ =18 d ^2-27 d +9 $
Differentiating with respect to $ d $
$ \Rightarrow 36 d -27=0 $
$ \Rightarrow d =\frac{3}{4}, \text { from }(1) a =\frac{-3}{2},( Z =\text { minimum }) $
Now, $ S _{ a }=\frac{ n }{2}\left(-3+( n -1) \frac{3}{4}\right)=0 $
$\Rightarrow n =5 $
Now,
$ n !-4 a _{ n ( n +2)}=120-4\left( a _{35}\right) $
$ =120-4( a +(35-1) d )$
$ =120-4\left(\frac{-3}{2}+34 \cdot\left(\frac{3}{4}\right)\right) $
$ =120-4\left(\frac{-6+102}{4}\right) $
$ =120-96=24$