JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Sequences and Series

Expanding the sum, ${\sum }_{r=1}^n(x+2r-2)(x+2r)={\sum }_{r=1}^n(x^2+(4r-2)x+(4r^2-4r))=n{x}^2+2n^{2}x+\frac{4n(n^2-1)}{3}$. Equating to $\frac{8n}{3}$ and dividing by $n$ gives $x^2+2nx+\frac{4n^2-12}{3}=0$. If the roots are two consecutive even integers their difference is $2$, so $(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta =4n^2-\frac{4(4n^2-12)}{3}=4$. This gives $12n^2-16n^2+48=12$, so $4n^2=36$ and $n=3$. For $n=3$ the equation becomes $x^2+6x+8=0$ with roots $-4$ and $-2$, two consecutive even integers. The required positive integer is $3$.