Let a₁, a₂, a₃, ... be an A.P. If a₇ = 3, the product a₁a₄ is minimum and the sum of its first n ter

Q: Wait, why do we reject the negative root or other values for $n$?

In an arithmetic progression, the number of terms $n$ must be a positive integer. When solving $S_n = 0$, we get a quadratic equation in $n$. One root is usually zero (which we reject because $n$ must be at least 1) and the other is our valid $n$.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure mathematics problem involving sequences and series. While physics problems often use A.P. for kinematics, here we rely on the algebraic properties of the sequence $a_n = a + (n-1)d$ and the optimization of the product.

Q: How do I know the product $P$ is a minimum and not a maximum?

Great question! Look at the coefficient of the $d^2$ term in your expanded quadratic $P(d) = 18d^2 - 27d + 9$. Since the coefficient $18$ is positive, the parabola opens upwards, which guarantees that the vertex is indeed a minimum.

Question

Let $a_1, a_2, a_3, \dots$ 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 :

Options:

(A) 24

(B) $\frac{33}{4}$

(C) $\frac{381}{4}$

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the start of a long, perfectly spaced path. Each step you take is exactly $d$ distance further than the last. This is the essence of an Arithmetic Progression (A.P.).

We are given an A.P. where the seventh term is $a_7 = 3$. Using the fundamental formula $a_n = a + (n-1)d$, we can immediately write:

$$a + 6d = 3$$

This is our anchor. It tells us that the first term $a$ is not independent; it is tethered to the common difference $d$ by the relationship:

$$a = 3 - 6d$$

### The Dance of the Product

The problem asks us to minimize the product of the first and fourth terms, $P = a_1 \cdot a_4$. Substituting our definitions, we have $a_1 = a$ and $a_4 = a + 3d$.

Thus, the product is $P = a(a + 3d)$. By substituting our anchor $a = 3 - 6d$, we transform this product into a function of a single variable:

$$P(d) = (3 - 6d)(3 - 6d + 3d) = (3 - 6d)(3 - 3d)$$

Expanding this, we get $P(d) = 9 - 9d - 18d + 18d^2$, which simplifies to the quadratic:

$$P(d) = 18d^2 - 27d + 9$$

### Finding the Minimum

We have a quadratic function $P(d) = 18d^2 - 27d + 9$. Since the coefficient of $d^2$ is positive, this parabola opens upwards, meaning its vertex is a minimum.

To find this minimum, we take the derivative with respect to $d$ and set it to zero:

$$\frac{dP}{dd} = 36d - 27 = 0$$

Solving this gives $d = \frac{27}{36} = \frac{3}{4}$. With $d$ in hand, we return to our anchor to find $a$:

$$a = 3 - 6\left(\frac{3}{4}\right) = 3 - \frac{9}{2} = -\frac{3}{2}$$

### The Sum Constraint

Now, we are told the sum of the first $n$ terms is zero. The sum formula is $S_n = \frac{n}{2}[2a + (n-1)d] = 0$.

Since $n$ cannot be zero, we must have $2a + (n-1)d = 0$. Substituting our values:

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

This simplifies to $-3 + (n-1)(\frac{3}{4}) = 0$. Solving for $n$, we find $(n-1)(\frac{3}{4}) = 3$, so $n-1 = 4$, which means **$n = 5$**.

### The Final Act

We are asked to calculate $n! - 4a_{n(n+2)}$. With $n=5$, this becomes $5! - 4a_{5(7)} = 120 - 4a_{35}$.

We calculate $a_{35}$ using the general term formula:

$$a_{35} = a + 34d = -\frac{3}{2} + 34\left(\frac{3}{4}\right) = -\frac{3}{2} + \frac{51}{2} = \frac{48}{2} = 24$$

Finally, we compute the result:

$$120 - 4(24) = 120 - 96 = 24$$

The journey from a simple sequence to this final, elegant result is a testament to the power of structured thinking. The final answer is **24**.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots$ 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 :

Select Answer:

Visualized Solution

Understanding the A.P. and $a_{7}$

Defining the Product $P = a_{1} a_{4}$

Substituting $a$ into the Product

Expanding the Quadratic Expression

Minimizing the Product using Calculus

Solving for Common Difference $d$

Finding the First Term $a$

Condition: Sum of $n$ terms is Zero

Substituting $a$ and $d$ into Sum Condition

Solving for $n$

Setting up the Target Expression

Calculating the 35th Term $a_{35}$

Final Calculation and Result

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Dance of the Product

Finding the Minimum

The Sum Constraint

The Final Act

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Select Answer:

Visualized Solution

Understanding the A.P. and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a7​

Defining the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=a1​a4​

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a into the Product

Expanding the Quadratic Expression

Minimizing the Product using Calculus

Solving for Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Finding the First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Condition: Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n terms is Zero

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