$\sum\limits_{r = 1}^n {(x + r – 1)(x + r) = 10n} $

$\sum\limits_{r = 1}^n {({x^2} + xr + (r – 1)x + {r^2} – r) = 10n} $

$ \Rightarrow \,\sum\limits_{r = 1}^n {({x^2} + (2r – 1)x + r(r – 1) = 10n} $

$\Rightarrow n x^{2}+\{1+3+5+\ldots .+(2 n-1)\} x+\{1.2+2.3+\ldots .+(n-1)n\}$

$=10 n$

$\Rightarrow \mathrm{nx}^{2}+\mathrm{n}^{2} \mathrm{x}+\frac{(\mathrm{n}-1) \mathrm{n}(\mathrm{n}+1)}{3}=10 \mathrm{n}$

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

Let $\alpha$ and $\alpha+1$ be its two solutions

( $\because$ it has two consequtive integral solutions)

$\Rightarrow \alpha+(\alpha+1)=-n$

$\Rightarrow \alpha=\frac{-n-1}{2}……….(1)$

Also $\alpha(\alpha+1)=\frac{n^{2}-31}{3}……..(2)$

Putting value of $(1)$ in $(2),$ we get

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

$ \Rightarrow {n^2} = 121$

$ \Rightarrow n = 11$