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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Sequences and Series

The positive integer n, for which the solutions of the equation...

The positive integer n, for which the solutions of the equation x(x+2)+(x+2)(x+4)+⋯+(x+2n−2)(x+2n)=8n3x(x+2)+(x+2)(x+4)+\cdots+(x+2 n-2)(x+2 n)=\frac{8 n}{3} are two consecutive even integers, is :

Options

  1. A3
  2. B12
  3. C9
  4. D6

Correct answer

A. 3

Step-by-step solution

The sum has nn terms: ∑k=1n(x+2k−2)(x+2k)\sum_{k=1}^{n}(x+2k-2)(x+2k).
Expanding and summing gives nx2+2n2x+4n(n2−1)3=8n3nx^2 + 2n^2x + \frac{4n(n^2-1)}{3} = \frac{8n}{3}.
Dividing by nn and simplifying: 3x2+6nx+4n2−12=03x^2 + 6nx + 4n^2 - 12 = 0.
Discriminant =36n2−12(4n2−12)=12(12−n2)= 36n^2 - 12(4n^2 - 12) = 12(12 - n^2).
Difference of roots =23(12−n2)3= \frac{2\sqrt{3(12-n^2)}}{3}.
For two consecutive even integers, this difference =2= 2.
36−3n2=3⇒n2=9⇒n=3\sqrt{36 - 3n^2} = 3 \Rightarrow n^2 = 9 \Rightarrow n = 3.
For n=3n = 3: x=−9±33=−2x = \frac{-9 \pm 3}{3} = -2 or −4-4, which are consecutive even integers.
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