Question

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 $n!-4a_{n(n+2)}$ is equal to :

• A. $\frac{381}{4}$
• B. 9
• C. $\frac{33}{4}$
• D. 24

Answer 1

Answer: (D)

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