Question

Let $a_1, a_2, a_3, \ldots$ be an AP 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. $\frac{381}{4}$
• B. 9
• C. $\frac{33}{4}$
• D. 24

Answer 1

Answer: (D)

24

Answer 2

The correct answer is (D) : 24
a+6d=3.......................(1)
Z=a(a+3d)
=(3−6d)(3−3d)
=18d2−27d+9
Differentiating with respect to d
⇒36d−27=0
⇒d=43​, from (1)a=2−3​,(Z= minimum )
Now, Sa​=2n​(−3+(n−1)43​)=0
⇒n=5
Now,
n!−4an(n+2)​=120−4(a35​)
=120−4(a+(35−1)d)
=120−4(2−3​+34⋅(43​))
=120−4(4−6+102​)
=120−96=24