Let a_1,a_2,a_3,…. 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!-4a_n(n+2) is equal to

We know the ${n}^{th}$nth term of an A.P. is given by,

$${a}_{n}=a+(n-1)d$$an​=a+(n−1)d

Given, ${a}_{7}=3$a7​=3

$$\Rightarrow a+6d=3$$⇒a+6d=3

$$\Rightarrow a=3-6d$$⇒a=3−6d

And, ${a}_{1}{a}_{4}=a(a+3d)$a1​a4​=a(a+3d)

$$=(3-6d)(3-3d)$$=(3−6d)(3−3d)

$$=18{d}^{2}-27d+9$$=18d2−27d+9

Given product $({a}_{1}{a}_{4})$(a1​a4​) is minimum then,

Let $f(d)=18{d}^{2}-27d+9$f(d)=18d2−27d+9

$${f}^{'}(d)=36d-27$$f′(d)=36d−27

Product to be minimum, ${f}^{'}(d)=0$f′(d)=0

$$\Rightarrow 36d-27=0$$⇒36d−27=0

$$\Rightarrow d=\frac{27}{36}=\frac{3}{4}$$⇒d=3627​=43​

So, $a=3-\frac{9}{2}=\frac{-3}{2}$a=3−29​=2−3​

Given, ${S}_{n}=0$Sn​=0

$${S}_{n}=\frac{n}{2}[2a+(n-1)d]=0$$Sn​=2n​[2a+(n−1)d]=0

$$-3+(n-1)\frac{3}{4}=0$$−3+(n−1)43​=0

$$\Rightarrow n=5$$⇒n=5

Now $n!-4{a}_{n(n+2)}=5!-4{a}_{35}$n!−4an(n+2)​=5!−4a35​

$$=120-4(a+34d)$$=120−4(a+34d)

$$=120-4(\frac{-3}{2}+34\times \frac{3}{4})$$=120−4(2−3​+34×43​)

$$=120+6-102=24$$=120+6−102=24