If, for a positive integer n , the quadratic equation, x x + 1 + x + 1 x + 2 + . .. + x + n - 1 ¯ x + n…

On simplifying we get the quadratic equations as
$\left(\underset{n \text{times}}{\underbrace{x^2+x^2+...+x^2}}\right)+\left[1+3+5+...+\left(2n-1\right)\right]x+\left[1.2+2.3+...+\left(n-1\right)n\right]=10n$
$n{x}^2+n^{2}x+\frac{n\left(n^2-1\right)}{3}=10n$

$x^2+nx+\frac{n^2-31}{3}=0$

Let, $\alpha , \beta$ are the roots of the above equation

$\therefore \alpha +\beta =-n, \alpha \beta =\frac{n^2-31}{3}$

Now, the difference of roots$\left|\alpha -\beta \right| =1$

$\Rightarrow {\left(\alpha -\beta \right)}^2=1$

$\Rightarrow {\left(\alpha +\beta \right)}^2-4\alpha \beta =1$

$\Rightarrow n^2-\frac{4}{3}\left(n^2-31\right)=1$

$\Rightarrow n^2=121$

$\Rightarrow n=11$