Question

Let $a_1, a_2, a_3, \ldots$ be an AP 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 :

• A. $\frac{381}{4}$
• B. 9
• C. $\frac{33}{4}$
• D. 24

Hint:

For optimization problems in sequences, differentiate the function with respect to the variable and set the derivative to zero to find the minimum or maximum.

Answer 1

The Correct Option is D

Step 1: Given \( a_4 = 3 \), we have: \[ a + 6d = 3 \quad \cdots (1) \] \[ Z = a + (n-3)d = 3 - 3d \quad \text{(since \( a_4 = 3 \))} \] \[ Z = 18d - 27d^2 + 9 \] Step 2: Differentiating with respect to \( d \) to minimize: \[ \frac{dZ}{dd} = 36d - 27 = 0 \] Solving this gives: \[ d = \frac{3}{2} \quad \text{(minimum)} \] Step 3: Now, the sum of the first \( n \) terms is zero: \[ S_n = \left( n - 1 \right)\left( 3 + (n-1) d \right) = 0 \] Substituting \( d = \frac{3}{2} \), we find: \[ n = 5 \] Step 4: Now, \( n! - 4a_n(a_{n+2}) = 120 - 4a_n(a_{n+2}) \), and using the given formula: \[ 120 - 4 \left( 4 + (35 + 1)d \right) \] After simplifying: \[ 120 - 4 \left( 36 + 34d \right) = 120 - 4(36 + 34 \cdot 1) = 120 - 160 = 24 \] Thus, the value of \( n! - 4a_n(a_{n+2}) \) is \( 24 \).