Q.1. Let , be the roots of the quadratic equation for some . Then is equal to :

Sol. $nx^2 + x(2 + 4 + 6 + ..... + 2n) + (1.3 + ..... + (n -1)(n + 1)) = 4n$ $nx^2 + n(n + 1)x + \frac{n(n-1)(2n + 5)}{6} = 4n$ $x^2 + (n + 1)x + \frac{(n-1)(2n + 5)}{6} - 4 = 0$ D must be a perfect square $D = (n+1)^2 - 4\left(\frac{(n-1)(2n+5)}{6} - 4\right) = \frac{n^2-1}{3}$ If make perfect square $\Rightarrow \frac{n^2-1}{3} = 16 \Rightarrow n = 7$ So Equation is $x^2 + 8x + \frac{8 \times 15}{6} - 4 = 0$ $x^2 + 8x + 15 = 0$ $x = -3, -5$ $\alpha = -5, \alpha + 2 = -3$ $a+n=7-5=2$

What is the correct answer for JEE Main Mathematics Q1?

The correct answer is C. Sol. $nx^2 + x(2 + 4 + 6 + ..... + 2n) + (1.3 + ..... + (n -1)(n + 1)) = 4n$ $nx^2 + n(n + 1)x + \frac{n(n-1)(2n + 5)}{6} = 4n$ $x^2 + (n + 1)x + \frac{(n-1)(2n + 5)}{6} - 4 = 0$ D must be a perfect square $D = (n+1)^2 - 4\left(\frac{(n-1)(2n+5)}{6} - 4\right) = \frac{n^2-1}{3}$ If make perfect square $\Rightarrow \frac{n^2-1}{3} = 16 \Rightarrow n = 7$ So Equation is $x^2 + 8x + \frac{8 \times 15

Q.1. Let , be the roots of the quadratic equation for some . Then is equal to :

The correct answer is C. Sol. $nx^2 + x(2 + 4 + 6 + ..... + 2n) + (1.3 + ..... + (n -1)(n + 1)) = 4n$ $nx^2 + n(n + 1)x + \frac{n(n-1)(2n + 5)}{6} = 4n$ $x^2 + (n + 1)x + \frac{(n-1)(2n + 5)}{6} - 4 = 0$ D must be a perfect square $D = (n+1)^2 - 4\left(\frac{(n-1)(2n+5)}{6} - 4\right) = \frac{n^2-1}{3}$ If make perfect square $\Rightarrow \frac{n^2-1}{3} = 16 \Rightarrow n = 7$ So Equation is $x^2 + 8x + \frac{8 \times 15

JEE Main 2026… PYQ · With Answer · Free

JEE Main•Mathematics•MCQs

Q.1. Let $a, a + 2, a \in Z$, be the roots of the quadratic equation $x(x + 2) + (x + 1)(x + 3) + (x + 2) (x + 4) +.....+ (x + n-1)(x + n + 1) = 4n$ for some $n \in N$. Then $(n + a)$ is equal to :

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