Polynomial Action

We can write the above as, $P(x) = \sum_{r=1}^n (x + r -1)(x +r) = 10n$

$P(x) = nx^2 + (\sum_{r=1}^n (2r - 1))x + \sum_{r=1}^n (r^2 -r) = 10n$

$P(x) = nx^2 + (n^2 + n -n) x + \dfrac {n(n+1)(n-1)}{3} - 10n= 0$

$P(x) = 3x^2 + 3nx + (n^2 - 31)= 0$

Now, $P(x)=0$ has two consecutive solutions $\alpha, \alpha + 1.$

We get two roots of $P(x)$ using quadratic formula as,

$\alpha = \dfrac {-3n + \sqrt{9n^2 - 12n^2 + 372}}{2(3)} , \alpha + 1= \dfrac {-3n - \sqrt{9n^2 -12n^2 + 372}}{2(3)}$

We know that, $(\alpha +1) - \alpha = 1 = \dfrac {-2\sqrt{372 - 3n^2}}{2\times 3} = 1 \Rightarrow 3n^2 = 372 - 9 = 363 \Rightarrow n^2 = 121 \Rightarrow n = \boxed{11}$