If, for a positive integer $$n$$, the quadratic equation,
$$x(x+1) + (x+1)(x+2) + \ldots + (x+\overline{n-1})(x+n) = 10n$$
has two consecutive integral solutions, then $$n$$ is equal to:

We begin with the expression

$$x(x+1)+(x+1)(x+2)+\ldots+(x+\overline{n-1})(x+n)=10n.$$

Observe that the $$k^{\text{th}}$$ term in this sum is $$(x+k)(x+k+1)$$ where $$k$$ runs from $$0$$ to $$n-1$$; hence there are exactly $$n$$ such terms. For any quantity $$a$$, we know the identity

$$(a)(a+1)=a^2+a.$$

Using this, every term becomes $$(x+k)^2+(x+k)$$. Therefore the entire left-hand side is

$$\sum_{k=0}^{n-1}\Big[(x+k)^2+(x+k)\Big].$$

We now split the summation and expand:

$$\sum_{k=0}^{n-1}(x+k)^2+\sum_{k=0}^{n-1}(x+k).$$

First we deal with the square term. Writing $$(x+k)^2=x^2+2xk+k^2$$, we find

$$\sum_{k=0}^{n-1}(x+k)^2 =\sum_{k=0}^{n-1}\big(x^2+2xk+k^2\big) =\underbrace{\sum_{k=0}^{n-1}x^2}_{=nx^2} +2x\underbrace{\sum_{k=0}^{n-1}k}_{=\dfrac{n(n-1)}2} +\underbrace{\sum_{k=0}^{n-1}k^2}_{=\dfrac{(n-1)n(2n-1)}6}.$$

Next we tackle the linear term:

$$\sum_{k=0}^{n-1}(x+k) =\sum_{k=0}^{n-1}x+\sum_{k=0}^{n-1}k =nx+\dfrac{n(n-1)}2.$$

Collecting all pieces, the entire sum equals

$$nx^2 +2x\cdot\dfrac{n(n-1)}2 +\dfrac{(n-1)n(2n-1)}6 +nx +\dfrac{n(n-1)}2.$$

The factor $$2$$ in the middle term cancels:

$$nx^2+n(n-1)x+\dfrac{(n-1)n(2n-1)}6+nx+\dfrac{n(n-1)}2.$$

Combining the two $$x$$-terms gives

$$nx^2+n^2x+\dfrac{(n-1)n(2n-1)}6+\dfrac{n(n-1)}2.$$

The two purely numerical fractions share the common factor $$n(n-1)$$, so we put them over the common denominator $$6$$:

$$\dfrac{(n-1)n(2n-1)}6+\dfrac{3(n-1)n}{6} =\dfrac{n(n-1)\big[(2n-1)+3\big]}6 =\dfrac{n(n-1)(2n+2)}6 =\dfrac{2n(n-1)(n+1)}6 =\dfrac{n(n-1)(n+1)}3.$$

Hence the given equation becomes

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

Because $$n\gt 0$$, we safely divide every term by $$n$$:

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

Noting that $$(n-1)(n+1)=n^2-1$$, we write

$$x^2+nx+\dfrac{n^2-1}{3}=10.$$

Moving the constant to one side gives the standard quadratic form in $$x$$:

$$x^2+nx+\frac{n^2-1}{3}-10=0,$$ $$x^2+nx+\frac{n^2-31}{3}=0.$$

Call the two integral roots $$r$$ and $$r+1$$ (because they are consecutive). For a quadratic $$x^2+Bx+C=0$$, the sum of roots is $$-B$$ and the product is $$C$$. Thus we have

$$r+(r+1)=2r+1=-n,\qquad r(r+1)=\frac{n^2-31}{3}.$$

From $$2r+1=-n$$ we obtain

$$r=-\frac{n+1}2.$$

We substitute this $$r$$ into the product condition:

$$r(r+1)=\left(-\frac{n+1}2\right)\left(-\frac{n+1}2+1\right) =\left(-\frac{n+1}2\right)\left(-\frac{n-1}2\right) =\frac{(n+1)(n-1)}4 =\frac{n^2-1}4.$$

Equating this with $$\dfrac{n^2-31}{3}$$ gives

$$\frac{n^2-1}4=\frac{n^2-31}3.$$

Cross-multiplying,

$$3(n^2-1)=4(n^2-31),$$ $$3n^2-3=4n^2-124.$$

Bringing all terms to one side,

$$0=4n^2-124-3n^2+3=n^2-121.$$

This yields

$$n^2=121\quad\Longrightarrow\quad n=11,$$

since $$n$$ is given to be a positive integer.

Hence, the correct answer is Option D.