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 $$|a| < 1, |b| < 1, |c| < 1, abc \neq 0$$, then

We are given $$x = \sum_{n=0}^{\infty} a^n$$, $$y = \sum_{n=0}^{\infty} b^n$$, $$z = \sum_{n=0}^{\infty} c^n$$, where $$a, b, c$$ are in A.P. with $$|a| < 1$$, $$|b| < 1$$, $$|c| < 1$$, and $$abc \neq 0$$.

Each sum is an infinite geometric series with first term $$1$$ and common ratio less than $$1$$ in absolute value. Using the formula $$\sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}$$ for $$|r| < 1$$:

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

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

Since $$a, b, c$$ are in A.P., we have $$2b = a + c$$, which gives $$a - b = b - c$$.

Now check if $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in A.P.: $$\frac{1}{y} - \frac{1}{x} = (1-b) - (1-a) = a - b$$ and $$\frac{1}{z} - \frac{1}{y} = (1-c) - (1-b) = b - c$$.

Since $$a - b = b - c$$, we have $$\frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y}$$, confirming that $$\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$$ are in A.P.

The answer is Option B.