Key Concepts and Formulas

• The $n^{th}$ term of an Arithmetic Progression (A.P.) is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
• The sum of the first $n$ terms of an A.P. is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.
• A quadratic function of the form $f(x) = Ax^2 + Bx + C$ has its minimum (if $A>0$) or maximum (if $A<0$) at $x = -\frac{B}{2A}$.

Step-by-Step Solution

Step 1: Express $a_1$ in terms of $d$ using the given $a_7=3$. We are given that $a_1, a_2, a_3, \ldots$ is an A.P. and $a_7 = 3$. Using the formula for the $n^{th}$ term, $a_n = a_1 + (n-1)d$, we have: $a_7 = a_1 + (7-1)d = a_1 + 6d$ Since $a_7 = 3$, we can write: $a_1 + 6d = 3$ From this, we can express $a_1$ in terms of $d$: $a_1 = 3 - 6d$

Step 2: Express the product $a_1 a_4$ as a function of $d$ and find the value of $d$ that minimizes it. We need to find the expression for $a_4$. Using the formula $a_n = a_1 + (n-1)d$: $a_4 = a_1 + (4-1)d = a_1 + 3d$ Now substitute the expression for $a_1$ from Step 1 into the expression for $a_4$: $a_4 = (3 - 6d) + 3d = 3 - 3d$ The product $a_1 a_4$ is: $P(d) = a_1 a_4 = (3 - 6d)(3 - 3d)$ Expand this expression: $P(d) = 9 - 9d - 18d + 18d^2$ $P(d) = 18d^2 - 27d + 9$ This is a quadratic function of $d$. Since the coefficient of $d^2$ (which is 18) is positive, the parabola opens upwards, and the function has a minimum value. The minimum occurs at the vertex, where $d = -\frac{b}{2a}$, with $a=18$ and $b=-27$. $d = -\frac{-27}{2(18)} = \frac{27}{36} = \frac{3}{4}$ This is the value of $d$ that minimizes the product $a_1 a_4$.

Step 3: Calculate $a_1$ using the minimized value of $d$. Substitute the value of $d = \frac{3}{4}$ back into the expression for $a_1$ from Step 1: $a_1 = 3 - 6d = 3 - 6\left(\frac{3}{4}\right) = 3 - \frac{18}{4} = 3 - \frac{9}{2}$ $a_1 = \frac{6}{2} - \frac{9}{2} = -\frac{3}{2}$

Step 4: Find the number of terms $n$ for which the sum of the first $n$ terms is zero. We are given that $S_n = 0$. Using the formula for the sum of the first $n$ terms, $S_n = \frac{n}{2}[2a_1 + (n-1)d]$: $S_n = \frac{n}{2}[2a_1 + (n-1)d] = 0$ Since $n$ must be a positive integer (number of terms), $n \neq 0$. Therefore, the term in the bracket must be zero: $2a_1 + (n-1)d = 0$ Substitute the values $a_1 = -\frac{3}{2}$ and $d = \frac{3}{4}$: $2\left(-\frac{3}{2}\right) + (n-1)\left(\frac{3}{4}\right) = 0$ $-3 + (n-1)\left(\frac{3}{4}\right) = 0$ $(n-1)\left(\frac{3}{4}\right) = 3$ Multiply both sides by $\frac{4}{3}$: $n-1 = 3 \cdot \frac{4}{3} = 4$ $n = 4 + 1 = 5$ So, the sum of the first 5 terms is zero.

Step 5: Calculate the value of $n! - 4a_{n(n+2)}$. We have $n=5$. First, calculate $n!$: $n! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ Next, calculate the index $n(n+2)$: $n(n+2) = 5(5+2) = 5(7) = 35$ Now, we need to find the $35^{th}$ term, $a_{35}$. Using the formula $a_n = a_1 + (n-1)d$: $a_{35} = a_1 + (35-1)d = a_1 + 34d$ Substitute the values $a_1 = -\frac{3}{2}$ and $d = \frac{3}{4}$: $a_{35} = -\frac{3}{2} + 34\left(\frac{3}{4}\right) = -\frac{3}{2} + \frac{102}{4}$ $a_{35} = -\frac{3}{2} + \frac{51}{2} = \frac{48}{2} = 24$ Finally, calculate the expression $n! - 4a_{n(n+2)}$: $n! - 4a_{n(n+2)} = 5! - 4a_{35} = 120 - 4(24)$ $120 - 96 = 24$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and signs. A small error can lead to a completely wrong answer.
• Confusing Indices: Carefully distinguish between the term number (like $n$ or 7) and the value of the term itself ($a_n$ or $a_7$).
• Minimization vs. Maximization: Remember that for a quadratic $Ax^2+Bx+C$, if $A>0$, the vertex gives a minimum. If $A<0$, it gives a maximum. In this problem, $A=18>0$, so we are indeed finding a minimum.

Summary

The problem requires us to first use the given information about the 7th term of an A.P. to establish a relationship between the first term ($a_1$) and the common difference ($d$). We then expressed the product $a_1 a_4$ as a quadratic in $d$ and found the value of $d$ that minimizes this product. Using this value of $d$, we determined $a_1$. Subsequently, we used the condition that the sum of the first $n$ terms is zero to find the value of $n$. Finally, we calculated the required expression $n! - 4a_{n(n+2)}$ using the determined values of $n$, $a_1$, and $d$.

The final answer is \boxed{24}, which corresponds to option (A).