If $$a_1, a_2, \ldots, a_n$$ are in H.P., then the expression $$a_1a_2 + a_2a_3 + \ldots + a_{n-1}a_n$$ is equal to

In a harmonic progression (H.P.) the reciprocals of the terms form an arithmetic progression (A.P.).
Hence write $$\frac{1}{a_k}=A+(k-1)D$$ for $$k=1,2,\ldots ,n$$, where $$A$$ is the first term and $$D$$ the common difference of the A.P.

This gives $$a_k=\frac{1}{A+(k-1)D}\quad\text{and}\quad a_{k+1}=\frac{1}{A+kD}$$.

Consider the product of two consecutive terms: $$a_k a_{k+1}= \frac{1}{\bigl(A+(k-1)D\bigr)\bigl(A+kD\bigr)}.$$

Now look at the difference of two successive reciprocals: $$\frac{1}{A+(k-1)D}-\frac{1}{A+kD}= \frac{D}{\bigl(A+(k-1)D\bigr)\bigl(A+kD\bigr)} = D\,a_k a_{k+1}.$$

Therefore $$a_k a_{k+1}= \frac{1}{D}\Bigl[\frac{1}{A+(k-1)D}-\frac{1}{A+kD}\Bigr].$$

Add these relations from $$k=1$$ to $$k=n-1$$: $$\sum_{k=1}^{n-1} a_k a_{k+1}= \frac{1}{D}\sum_{k=1}^{n-1}\Bigl[\frac{1}{A+(k-1)D}-\frac{1}{A+kD}\Bigr].$$

The series on the right telescopes - every intermediate term cancels - leaving $$\sum_{k=1}^{n-1} a_k a_{k+1}= \frac{1}{D}\Bigl(\frac{1}{A}-\frac{1}{A+(n-1)D}\Bigr).$$

Since $$\frac{1}{A}=a_1$$ and $$\frac{1}{A+(n-1)D}=a_n$$, the sum becomes $$S=\frac{1}{D}(a_1-a_n).$$

We still need $$\frac{1}{D}$$. From the A.P. of reciprocals, $$\frac{1}{a_n}-\frac{1}{a_1}=(n-1)D \;\Longrightarrow\; D=\frac{\frac{1}{a_n}-\frac{1}{a_1}}{n-1}= \frac{a_1-a_n}{(n-1)a_1a_n}.$$ Hence $$\frac{1}{D}= \frac{(n-1)a_1a_n}{a_1-a_n}.$$

Substituting this in $$S$$ yields $$S=\frac{1}{D}(a_1-a_n)=\Bigl[\frac{(n-1)a_1a_n}{a_1-a_n}\Bigr](a_1-a_n)=(n-1)a_1a_n.$$ Thus $$\boxed{a_1a_2+a_2a_3+\ldots+a_{n-1}a_n=(n-1)a_1a_n}.$$

Therefore the correct choice is:
Option D which is: $$(n - 1)a_1a_n$$