JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Minimizing a product of terms and finding parameters of an Arithmetic Progression

Q: 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).

Let a_1, a_2, a_3, … 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:

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A9B33/4C381/4D24

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 $\frac{n}{2} [2 a_{1} + (n - 1) d] = 0$ carefully: with $a_{1} = - \frac{3}{4}$ and $d = \frac{3}{8}$ , $2 (- \frac{3}{4}) + (n - 1) \frac{3}{8} = 0 ⟹ n - 1 = 4$ , so $n = 5$ .

B · 33/4

The student computed only the term $a_{n (n + 2)}$ or slipped while combining $n!$ with $4 a_{n (n + 2)}$ . Evaluate each term completely: $n! = 5! = 120$ , $n (n + 2) = 35$ , $a_{35} = a_{1} + 34 d = 12$ , and compute $120 - 4 (12) = 72$ , or avoid arithmetic slips in evaluating the full required expression.

C · 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 $a_{1} = 3 - 6 d$ and $a_{4} = 3 - 3 d$ . Their product is $(3 - 6 d) (3 - 3 d) = 9 (2 d - 1) (d - 1) = 9 (2 d^{2} - 3 d + 1)$ . The minimum occurs at $d = - \frac{- 3}{2 (2)} = \frac{3}{4}$ or by differentiation, giving $d = \frac{3}{8}$ .

D · 24

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 $4 a_{35}$ was evaluated.

Reviewed route

Solution

StepWorking

01Given

For an A.P. with first term $a_{1}$ and common difference $d$ : 1. $a_{7} = a_{1} + 6 d = 3$ 2. The product $P = a_{1} a_{4}$ is minimized. 3. $S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d] = 0$ for $n > 0$ .

02Goal

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

03Approach

Express $a_{1}$ in terms of $d$ using $a_{1} = 3 - 6 d$ . Express $P = a_{1} a_{4} = a_{1} (a_{1} + 3 d)$ 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 $n! - 4 a_{n (n + 2)}$ .

04Execute

Substitute $a_{1} = 3 - 6 d$ into $a_{4} = a_{1} + 3 d = 3 - 3 d$ . The product is: $f (d) = (3 - 6 d) (3 - 3 d) = 18 d^{2} - 27 d + 9$ 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 $a_{1}$ : $a_{1} = 3 - 6 (\frac{3}{4}) = 3 - \frac{9}{2} = - \frac{3}{2}$

06Execute

Using $S_{n} = 0$ where $n \neq 0$ : $2 a_{1} + (n - 1) d = 0 ⟹ 2 (- \frac{3}{2}) + (n - 1) (\frac{3}{4}) = 0$ $- 3 + \frac{3}{4} (n - 1) = 0 ⟹ n - 1 = 4 ⟹ n = 5$

07Execute

Evaluate the target expression $n! - 4 a_{n (n + 2)}$ for $n = 5$ : $n (n + 2) = 5 (7) = 35$ $a_{35} = a_{1} + 34 d = - \frac{3}{2} + 34 (\frac{3}{4}) = - \frac{3}{2} + \frac{51}{2} = \frac{48}{2} = 24$ Now compute $5! - 4 a_{35}$ : $5! - 4 (24) = 120 - 96 = 24$

✓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 $a_{7} = a_{1} + 6 d = 3$ , express the product $P (d) = a_{1} a_{4}$ as a function of the common difference $d$ .
$P (d) = (3 - 6 d) (3 - 3 d) = 18 d^{2} - 27 d + 9$
For what value of $d$ is $P (d) = 18 d^{2} - 27 d + 9$ 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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why can we minimize

P

using

- b / (2 a)

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 $d = - b / (2 a)$ .

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 $\frac{n}{2} [2 a_{1} + (n - 1) d] = 0$ carefully: with $a_{1} = - \frac{3}{4}$ and $d = \frac{3}{8}$ , $2 (- \frac{3}{4}) + (n - 1) \frac{3}{8} = 0 ⟹ n - 1 = 4$ , so $n = 5$ .
B: 33/4 - The student computed only the term $a_{n (n + 2)}$ or slipped while combining $n!$ with $4 a_{n (n + 2)}$ . Evaluate each term completely: $n! = 5! = 120$ , $n (n + 2) = 35$ , $a_{35} = a_{1} + 34 d = 12$ , and compute $120 - 4 (12) = 72$ , or avoid arithmetic slips in evaluating the full required expression.
C: 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 $a_{1} = 3 - 6 d$ and $a_{4} = 3 - 3 d$ . Their product is $(3 - 6 d) (3 - 3 d) = 9 (2 d - 1) (d - 1) = 9 (2 d^{2} - 3 d + 1)$ . The minimum occurs at $d = - \frac{- 3}{2 (2)} = \frac{3}{4}$ or by differentiation, giving $d = \frac{3}{8}$ .
D · correct: 24 - 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 $4 a_{35}$ was evaluated.

Step-by-step solution

Given: For an A.P. with first term $a_{1}$ and common difference $d$ : 1. $a_{7} = a_{1} + 6 d = 3$ 2. The product $P = a_{1} a_{4}$ is minimized. 3. $S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d] = 0$ for $n > 0$ .
Goal: Find the value of $n! - 4 a_{n (n + 2)}$ .
Approach: Express $a_{1}$ in terms of $d$ using $a_{1} = 3 - 6 d$ . Express $P = a_{1} a_{4} = a_{1} (a_{1} + 3 d)$ 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 $n! - 4 a_{n (n + 2)}$ .
Execute: Substitute $a_{1} = 3 - 6 d$ into $a_{4} = a_{1} + 3 d = 3 - 3 d$ . The product is: $f (d) = (3 - 6 d) (3 - 3 d) = 18 d^{2} - 27 d + 9$ 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 $a_{1}$ : $a_{1} = 3 - 6 (\frac{3}{4}) = 3 - \frac{9}{2} = - \frac{3}{2}$
Execute: Using $S_{n} = 0$ where $n \neq 0$ : $2 a_{1} + (n - 1) d = 0 ⟹ 2 (- \frac{3}{2}) + (n - 1) (\frac{3}{4}) = 0$ $- 3 + \frac{3}{4} (n - 1) = 0 ⟹ n - 1 = 4 ⟹ n = 5$
Execute: Evaluate the target expression $n! - 4 a_{n (n + 2)}$ for $n = 5$ : $a_{1} = 3 - 6 (\frac{3}{4}) = 3 - \frac{9}{2} = - \frac{3}{2}$ $a_{35} = a_{1} + 34 d = - \frac{3}{2} + 34 (\frac{3}{4}) = - \frac{3}{2} + \frac{51}{2} = \frac{48}{2} = 24$ Now compute $5! - 4 a_{35}$ : $5! - 4 (24) = 120 - 96 = 24$
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.

Minimizing a product of terms and finding parameters of an Arithmetic Progression

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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