Class JEE Mathematics ALL Q #1195
KNOWLEDGE BASED
APPLY
4 Marks 2026 JEE Main 2026 (Online) 22 January Morning Shift MCQ SINGLE
If sum of first 4 terms of an A.P. is 6 and sum of first 6 terms is 4, then sum of first 12 terms of an A.P. is :
(A) -22
(B) -21
(C) -23
(D) -24
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AI Tutor Explanation

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Step-by-Step Solution

Let the first term of the A.P. be 'a' and the common difference be 'd'.

The sum of the first n terms of an A.P. is given by: Sn = (n/2) * [2a + (n-1)d]

Given, S4 = 6. Therefore, (4/2) * [2a + (4-1)d] = 6 => 2 * (2a + 3d) = 6 => 2a + 3d = 3 ----(1)

Also, S6 = 4. Therefore, (6/2) * [2a + (6-1)d] = 4 => 3 * (2a + 5d) = 4 => 2a + 5d = 4/3 ----(2)

Subtracting equation (1) from equation (2), we get: (2a + 5d) - (2a + 3d) = (4/3) - 3 => 2d = (4 - 9)/3 => 2d = -5/3 => d = -5/6

Substituting the value of d in equation (1), we get: 2a + 3*(-5/6) = 3 => 2a - 5/2 = 3 => 2a = 3 + 5/2 => 2a = 11/2 => a = 11/4

Now, we need to find S12. S12 = (12/2) * [2a + (12-1)d] => S12 = 6 * [2*(11/4) + 11*(-5/6)] => S12 = 6 * [11/2 - 55/6] => S12 = 6 * [(33 - 55)/6] => S12 = 6 * (-22/6) => S12 = -22

Correct Answer: -22

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the formulas for the sum of an arithmetic progression to solve for unknown variables and then calculate the sum of the first 12 terms.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure: using the given information to set up equations, solving for the first term and common difference, and then using these values to calculate the sum of the first 12 terms.
Syllabus Audit: In the context of JEE, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of the formulas related to arithmetic progressions, a standard topic in the JEE syllabus.

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