If $$x = \sum_{n=0}^\infty a^n, y = \sum_{n=0}^\infty b^n, z = \sum_{n=0}^\infty c^n$$ where $$a, b, c$$ are in A.P. and $$|a| < 1, |b| < 1, |c| < 1$$, then $$x, y, z$$ are in

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.