JEE Main 2026 Jan 21 Shift 2, Mathematics Q1: Quadratic Equations

JEE Main2026Jan 21, Shift 2Mathematics

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

$x (x + 2) + (x + 2) (x + 4) + \dots + (x + 2 n - 2) (x + 2 n) = \frac{8 n}{3}$

are two consecutive even integers, is :-

A
3
B
6
C
12
D
9

Solution

Rewrite the sum as $\sum_{r = 1}^{n} (x + 2 r - 2) (x + 2 r) = \frac{8 n}{3}$ .

Expanding each term and summing the standard $\sum r$ and $\sum r^{2}$ pieces:

$n x^{2} + 2 x \sum_{r = 1}^{n} (2 r - 1) + 4 \sum_{r = 1}^{n} r (r - 1) = \frac{8 n}{3}$

$n x^{2} + 2 x \cdot n^{2} + \frac{4 n ( n ^{2} - 1 )}{3} - \frac{8 n}{3} = 0$

Dividing by $n$ gives a quadratic in $x$ : $x^{2} + 2 n x + \frac{4 ( n ^{2} - 1 )}{3} - \frac{8}{3} = 0$ .

For two consecutive even-integer roots $α, β$ we need $∣ α - β ∣ = 2$ , i.e. $\frac{D}{∣ a ∣} = 2 \Rightarrow D = 4$ .

Here $D = 4 n^{2} - 4 (\frac{4 ( n ^{2} - 1 )}{3} - \frac{8}{3}) = 4$ , which simplifies to $n^{2} - \frac{4 n ^{2}}{3} = - 3$ , giving $n^{2} = 9$ , so $n = 3$ .

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JEE Main 2026 Jan 21 Shift 2, Mathematics Q1: Quadratic Equations

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