Sequences And Series Question 16
Question 16 - 31 January - Shift 2
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:
(1) 24
(2) $\frac{33}{4}$
(3) $\frac{381}{4}$
(4) 9
Show Answer
Answer: (1)
Solution:
Formula: General Term of an AP and Sum of $\mathbf{n}$ terms of an AP
$a+6 d=3$,
$Z=a(a+3 d)$
$=(3-6 d)(3-3 d)$
$=18 d^{2}-27 d+9$
Differentiating with respect to $d$
$\Rightarrow 36 d-27=0$
$\Rightarrow d=\frac{3}{4}$, from $(1) a=\frac{-3}{2},(Z=$ minimum $)$
Now, $S_n=\frac{n}{2}(-3+(n-1) \frac{3}{4})=0$
$\Rightarrow n=5$
Now,
$ \begin{aligned} & n !-4 a _{n(n+2)}=120-4(a _{35}) \\ & =120-4(a+(35-1) d) \\ & =120-4(\frac{-3}{2}+34 \cdot(\frac{3}{4})) \\ & =120-4(\frac{-6+102}{4}) \\ & =120-96=24 \end{aligned} $
