JEE Main 2026 — 21 January, Evening Shift — Question 1

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

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

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

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

$\mathrm{x}^{2}+2 \mathrm{nx}+\frac{4\left(\mathrm{n}^{2}-1\right)}{3}-\frac{8}{3}=0<_{\beta}^{\alpha}$

$\because|\alpha-\beta|=2 \Rightarrow \frac{\sqrt{\mathrm{D}}}{|\mathrm{a}|}=2 \Rightarrow \mathrm{D}=4$

$\Rightarrow 4 \mathrm{n}^{2}-4\left(4 \frac{\left(\mathrm{n}^{2}-1\right)}{3}-\frac{8}{3}\right)=4$

$\Rightarrow \mathrm{n}^{2}-\frac{4 \mathrm{n}^{2}}{3}=-3$

$\Rightarrow \mathrm{n}^{2}=9$ $\Rightarrow \mathrm{n}=3$