If, for a positive integer n, the quadratic equation, x(x+1)+(x+1)(x+2)+...+(x+ n-1)(x+n)=10n has two consecutive integral solutions, then n is equal…

On simplifying we get the quadratic equations as

$$(\underset{n\text{times}}{\underset{⏟}{{x}^{2}+{x}^{2}+...+{x}^{2}}})+[1+3+5+...+(2n-1)]x+[1.2+2.3+...+(n-1)n]=10n$$(ntimes⏟x2+x2+...+x2​​)+[1+3+5+...+(2n−1)]x+[1.2+2.3+...+(n−1)n]=10n

$$n{x}^{2}+{n}^{2}x+\frac{n({n}^{2}-1)}{3}=10n$$nx2+n2x+3n(n2−1)​=10n

$${x}^{2}+nx+\frac{{n}^{2}-31}{3}=0$$x2+nx+3n2−31​=0

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

$$\therefore \alpha +\beta =-n,\alpha \beta =\frac{{n}^{2}-31}{3}$$∴α+β=−n,αβ=3n2−31​

Now, the difference of roots$|\alpha -\beta | =1$∣α−β∣=1

$$\Rightarrow {(\alpha -\beta )}^{2}=1$$⇒(α−β)2=1

$$\Rightarrow {(\alpha +\beta )}^{2}-4\alpha \beta =1$$⇒(α+β)2−4αβ=1

$$\Rightarrow {n}^{2}-\frac{4}{3}({n}^{2}-31)=1$$⇒n2−34​(n2−31)=1

$$\Rightarrow {n}^{2}=121$$⇒n2=121

$$\Rightarrow n=11$$⇒n=11