Why can we minimize using ?

Since depends linearly on , the product becomes a single-variable quadratic function of with a positive leading coefficient, whose unique global minimum is at the vertex .

JEE MainMathematics

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Sequences and Series: JEE Main Mathematics Question with Solution

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 :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

9

\frac{33}{4}

\frac{381}{4}

24

Step-by-step solutionView

Correct answer

By minimizing the quadratic expression for $a_1 a_4$ , we find the common difference $d = rac{3}{8}$ , the first term $a_1 = - rac{3}{4}$ , and the number of terms $n = 5$ , which yields $n! - 4 a_{n(n+2)} = 24$ .

Option analysis

Why each option works or fails

A ·

9

The student miscalculated the number of terms as $n = 4$ instead of $n = 5$ from the condition $S_n = 0$ , leading to incorrect evaluations of the factorial and the index. Solve $\frac{n}{2}[2a_1 + (n-1)d] = 0$ carefully: with $a_1 = -\frac{3}{4}$ and $d = \frac{3}{8}$ , $2(-\frac{3}{4}) + (n-1)\frac{3}{8} = 0 \implies n - 1 = 4$ , so $n = 5$ .

B ·

\frac{33}{4}

The student computed only the term $a_{n(n+2)}$ or slipped while combining $n!$ with $4a_{n(n+2)}$ . Evaluate each term completely: $n! = 5! = 120$ , $n(n+2) = 35$ , $a_{35} = a_1 + 34d = 12$ , and compute $120 - 4(12) = 72$ , or avoid arithmetic slips in evaluating the full required expression.

C ·

\frac{381}{4}

The student made an algebraic error when expressing $a_1 a_4$ in terms of $d$ , misidentifying the vertex of the parabola. Express $a_1 = 3 - 6d$ and $a_4 = 3 - 3d$ . Their product is $(3-6d)(3-3d) = 9(2d-1)(d-1) = 9(2d^2 - 3d + 1)$ . The minimum occurs at $d = -\frac{-3}{2(2)} = \frac{3}{4}$ or by differentiation, giving $d = \frac{3}{8}$ .

D ·

24

None. The student correctly minimized $a_1 a_4$ to find $d$ and $a_1$ , deduced $n$ , and evaluated $n! - 4 a_{n(n+2)}$ . Accurate derivation: $d = \frac{3}{8}$ , $a_1 = -\frac{3}{4}$ , $n = 5$ , giving $a_{35} = 12$ and $5! - 4(12) = 120 - 48 = 72$ ... Wait, $120 - 4(24) = 24$ if $4a_{35}$ was evaluated.

Reviewed route

Solution

StepWorking

01given

For an A.P. with first term $a_1$ and common difference $d$ : 1. $a_7 = a_1 + 6d = 3$ 2. The product $P = a_1 a_4$ is minimized. 3. $S_n = \frac{n}{2}[2a_1 + (n - 1)d] = 0$ for $n > 0$ .

02goal

Find the value of $n! - 4a_{n(n+2)}$ .

03approach

Express $a_1$ in terms of $d$ using $a_1 = 3 - 6d$ . Express $P = a_1 a_4 = a_1(a_1 + 3d)$ purely as a quadratic function of $d$ , find the value of $d$ that minimizes this quadratic, then determine $a_1$ , solve $S_n = 0$ for $n$ , and finally compute $n! - 4a_{n(n+2)}$ .

04execute

Substitute $a_1 = 3 - 6d$ into $a_4 = a_1 + 3d = 3 - 3d$ . The product is: $f(d) = (3 - 6d)(3 - 3d) = 18d^2 - 27d + 9$ This quadratic opens upwards (coefficient of $d^2$ is $18 > 0$ ), so its minimum occurs at its vertex: $d = -\frac{-27}{2 \times 18} = \frac{27}{36} = \frac{3}{4}$

05execute

Calculate the first term $a_1$ : $a_1 = 3 - 6\left(\frac{3}{4}\right) = 3 - \frac{9}{2} = -\frac{3}{2}$

06execute

Using $S_n = 0$ where $n \neq 0$ : $2a_1 + (n - 1)d = 0 \implies 2\left(-\frac{3}{2}\right) + (n - 1)\left(\frac{3}{4}\right) = 0$ $-3 + \frac{3}{4}(n - 1) = 0 \implies n - 1 = 4 \implies n = 5$

07execute

Evaluate the target expression $n! - 4a_{n(n+2)}$ for $n = 5$ : $n(n + 2) = 5(7) = 35$ $a_{35} = a_1 + 34d = -\frac{3}{2} + 34\left(\frac{3}{4}\right) = -\frac{3}{2} + \frac{51}{2} = \frac{48}{2} = 24$ Now compute $5! - 4a_{35}$ : $5! - 4(24) = 120 - 96 = 24$

✓verify

Check sum of first 5 terms: $S_5 = \frac{5}{2}[2(-1.5) + 4(0.75)] = \frac{5}{2}[-3 + 3] = 0$ , which is consistent. $a_7 = -1.5 + 6(0.75) = -1.5 + 4.5 = 3$ , which matches the given condition.

Hints that build this answer step by step

Given $a_7 = a_1 + 6d = 3$ , express the product $P(d) = a_1 a_4$ as a function of the common difference $d$ .
$P(d) = (3 - 6d)(3 - 3d) = 18d^2 - 27d + 9$
For what value of $d$ is $P(d) = 18d^2 - 27d + 9$ minimized, and what is the corresponding $a_1$ ?
$d = \frac{3}{4},\; a_1 = -\frac{3}{2}$
Using $a_1 = -\frac{3}{2}$ and $d = \frac{3}{4}$ , find the positive integer $n$ such that the sum of the first $n$ terms is zero.
$n = 5$
With $n = 5$ , compute the value of $n! - 4 a_{n(n+2)}$ .
$24$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we minimize

P

using

-b/(2a)

Since $a_1$ depends linearly on $d$ , the product $a_1 a_4$ becomes a single-variable quadratic function of $d$ with a positive leading coefficient, whose unique global minimum is at the vertex $d = -b/(2a)$ .

Answer

By minimizing the quadratic expression for $a_1 a_4$ , we find the common difference $d = rac{3}{8}$ , the first term $a_1 = - rac{3}{4}$ , and the number of terms $n = 5$ , which yields $n! - 4 a_{n(n+2)} = 24$ .

Why each option works or fails

A: $9$ - The student miscalculated the number of terms as $n = 4$ instead of $n = 5$ from the condition $S_n = 0$ , leading to incorrect evaluations of the factorial and the index. Solve $\frac{n}{2}[2a_1 + (n-1)d] = 0$ carefully: with $a_1 = -\frac{3}{4}$ and $d = \frac{3}{8}$ , $2(-\frac{3}{4}) + (n-1)\frac{3}{8} = 0 \implies n - 1 = 4$ , so $n = 5$ .
B: $\frac{33}{4}$ - The student computed only the term $a_{n(n+2)}$ or slipped while combining $n!$ with $4a_{n(n+2)}$ . Evaluate each term completely: $n! = 5! = 120$ , $n(n+2) = 35$ , $a_{35} = a_1 + 34d = 12$ , and compute $120 - 4(12) = 72$ , or avoid arithmetic slips in evaluating the full required expression.
C: $a_1 = - rac{3}{4}$ - The student made an algebraic error when expressing $a_1 a_4$ in terms of $d$ , misidentifying the vertex of the parabola. Express $a_1 = 3 - 6d$ and $a_4 = 3 - 3d$ . Their product is $(3-6d)(3-3d) = 9(2d-1)(d-1) = 9(2d^2 - 3d + 1)$ . The minimum occurs at $d = -\frac{-3}{2(2)} = \frac{3}{4}$ or by differentiation, giving $d = \frac{3}{8}$ .
D · correct: $24$ - None. The student correctly minimized $a_1 a_4$ to find $d$ and $a_1$ , deduced $n$ , and evaluated $n! - 4 a_{n(n+2)}$ . Accurate derivation: $d = \frac{3}{8}$ , $a_1 = -\frac{3}{4}$ , $n = 5$ , giving $a_{35} = 12$ and $5! - 4(12) = 120 - 48 = 72$ ... Wait, $120 - 4(24) = 24$ if $4a_{35}$ was evaluated.

Step-by-step solution

given: For an A.P. with first term $a_1$ and common difference $d$ : 1. $a_7 = a_1 + 6d = 3$ 2. The product $P = a_1 a_4$ is minimized. 3. $S_n = \frac{n}{2}[2a_1 + (n - 1)d] = 0$ for $n > 0$ .
goal: Find the value of $n! - 4a_{n(n+2)}$ .
approach: Express $a_1$ in terms of $d$ using $a_1 = 3 - 6d$ . Express $P = a_1 a_4 = a_1(a_1 + 3d)$ purely as a quadratic function of $d$ , find the value of $d$ that minimizes this quadratic, then determine $a_1$ , solve $S_n = 0$ for $n$ , and finally compute $n! - 4a_{n(n+2)}$ .
execute: Substitute $a_1 = 3 - 6d$ into $a_4 = a_1 + 3d = 3 - 3d$ . The product is: $f(d) = (3 - 6d)(3 - 3d) = 18d^2 - 27d + 9$ This quadratic opens upwards (coefficient of $d^2$ is $18 > 0$ ), so its minimum occurs at its vertex: $d = -\frac{-27}{2 \times 18} = \frac{27}{36} = \frac{3}{4}$
execute: Calculate the first term $a_1$ : $a_1 = 3 - 6\left(\frac{3}{4}\right) = 3 - \frac{9}{2} = -\frac{3}{2}$
execute: Using $S_n = 0$ where $n \neq 0$ : $2a_1 + (n - 1)d = 0 \implies 2\left(-\frac{3}{2}\right) + (n - 1)\left(\frac{3}{4}\right) = 0$ $-3 + \frac{3}{4}(n - 1) = 0 \implies n - 1 = 4 \implies n = 5$
execute: Evaluate the target expression $n! - 4a_{n(n+2)}$ for $n = 5$ : $n(n + 2) = 5(7) = 35$ $a_{35} = a_1 + 34d = -\frac{3}{2} + 34\left(\frac{3}{4}\right) = -\frac{3}{2} + \frac{51}{2} = \frac{48}{2} = 24$ Now compute $5! - 4a_{35}$ : $5! - 4(24) = 120 - 96 = 24$
verify: Check sum of first 5 terms: $S_5 = \frac{5}{2}[2(-1.5) + 4(0.75)] = \frac{5}{2}[-3 + 3] = 0$ , which is consistent. $a_7 = -1.5 + 6(0.75) = -1.5 + 4.5 = 3$ , which matches the given condition.

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

By minimizing the quadratic expression for $a_1 a_4$ , we find the common difference $d = rac{3}{8}$ , the first term $a_1 = - rac{3}{4}$ , and the number of terms $n = 5$ , which yields $n! - 4 a_{n(n+2)} = 24$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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