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 $a_1a_4$ is minimum, and the sum of its first $n$ terms is zero, then $n! - 4a_{n(n+2)}$ is equal to:

Accepted Answer

$a_7=a+6d=3 \Rightarrow a=3-6d$. $Z=a_1a_4 = a(a+3d) = (3-6d)(3-3d) = 18d^2-27d+9$.

Minimizing: $Z'(d)=36d-27=0 \Rightarrow d=\dfrac34$, giving $a=-\dfrac32$.

$S_n=0$: $\dfrac{n}{2}\left[-3+(n-1)\dfrac34\right]=0 \Rightarrow n=5$.

$$n(n+2) = 5\times7=35, \qquad a_{35} = -\frac32+34\left(\frac34\right)=24.$$

$n!-4a_{35} = 120-4(24) = 120-96 = 24.$

Solutions · 1