Harmonic Progression

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: $$(n - 1)a_1a_n$$

The given infinite geometric series converge because $$|a| \lt 1,\; |b| \lt 1,\; |c| \lt 1$$.

For an infinite G.P. with first term $$1$$ and common ratio $$r$$ (where $$|r| \lt 1$$), the sum is $$\displaystyle\frac{1}{1-r}$$. Applying this:

$$x=\sum_{n=0}^{\infty}a^{n}=\frac{1}{1-a},\qquad y=\sum_{n=0}^{\infty}b^{n}=\frac{1}{1-b},\qquad z=\sum_{n=0}^{\infty}c^{n}=\frac{1}{1-c}.$$

To check whether $$x,y,z$$ form an H.P., look at their reciprocals. We obtain

$$\frac{1}{x}=1-a,\qquad \frac{1}{y}=1-b,\qquad \frac{1}{z}=1-c.$$

Given that $$a,b,c$$ are in arithmetic progression, there exists a common difference $$d$$ such that $$b-a=d,\quad c-b=d.$$

Then $$(1-b)-(1-a)=a-b=-d,\qquad (1-c)-(1-b)=b-c=-d,$$ so $$1-a,\;1-b,\;1-c$$ are also in arithmetic progression.

Thus $$\displaystyle\frac{1}{x},\;\frac{1}{y},\;\frac{1}{z}$$ form an A.P. Therefore, $$x,y,z$$ form a harmonic progression (H.P.).

Option D which is: H.P.