Let a_1, a_2, a_3, be an A.P.

Question

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 :

Accepted Answer

$a_{7}=3 \Rightarrow a+6 d=3 \Rightarrow a=3-6 d$

$$
\begin{aligned}
& a_{1} \cdot a_{4}=a(a+3 d) \\
& \Rightarrow(3-6 d)(3-6 d+3 d) \\
& \Rightarrow 3(1-2 d) 3(1-d) \\
& \Rightarrow 9\left(2 d^{2}-3 d+1\right)
\end{aligned}
$$

Let $f(d)=2 d^{2}-3 d+1$

$f^{\prime}(d)=4 d-3 \Rightarrow d=\frac{3}{4}$

$\therefore a=3-6 \cdot \frac{3}{4}=3-\frac{9}{2}=-\frac{3}{2}$

$S_{n}=0$

$\frac{n}{2}(29+(n-1) d)=0$

$\Rightarrow 2 \cdot\left(-\frac{3}{2}\right)+(n-1)\left(\frac{3}{4}\right)=0$

$\Rightarrow \quad 3=\frac{3}{4}(n-1)$

$\Rightarrow n=5$

Now, $n !-4 \cdot a_{n(n+2)}$

$$
\begin{aligned}
& =5 !-4 \cdot a_{35} \\
& =120-4\left(-\frac{3}{2}+34 \cdot \frac{3}{4}\right) \\
& =120-(-6+102) \\
& =120-(96) \\
& =24
\end{aligned}
$$

Solutions · 1