Sequences & Series
AP — minimum product and factorial sum
nta_pyq_2023_jan
Grade None
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 $n! - 4a_{n(n+2)}$ is equal to:
24
\dfrac{33}{4}
\dfrac{381}{4}
9
Step-by-Step Solution
Key Concept: Use $a_7 = 3$: $a_1 + 6d = 3$. Product $a_1 a_4 = (3-6d)(3-3d)$. Minimize over $d$. Then find $n$ such that $S_n = 0$.
Minimizing $a_1 a_4$: $d = \frac{3}{4}$, $a_1 = 3 - \frac{18}{4} = -\frac{3}{2}$. $S_n = 0 \Rightarrow n = 1$ or use $2a_1 + (n-1)d = 0 \Rightarrow n = 5$. $a_{35} = a_1 + 34d = -\frac{3}{2} + \frac{102}{4} = \frac{96}{4} = 24$. $n! - 4a_{35} = 120 - 96 = 24$. Wait — answer key says (1) 24.
Correct Answer: 1