Key Concepts and Formulas

• Sum of an Infinite Geometric Progression (G.P.): If a G.P. has first term $A$ and common ratio $R$, and $|R| < 1$, its sum to infinity is given by $S_\infty = \frac{A}{1-R}$.
• Arithmetic Progression (A.P.): A sequence $p, q, r, \dots$ is in A.P. if the difference between consecutive terms is constant (i.e., $q-p = r-q$). An important property is that if $p, q, r$ are in A.P., then $p+k, q+k, r+k$ are also in A.P. for any constant $k$, and $kp, kq, kr$ are also in A.P. for any non-zero constant $k$.
• Harmonic Progression (H.P.): A sequence of non-zero numbers $u, v, w, \dots$ is in H.P. if their reciprocals $\frac{1}{u}, \frac{1}{v}, \frac{1}{w}, \dots$ are in A.P.

Step-by-Step Solution

Step 1: Expressing x, y, and z using the sum of infinite G.P.

We are given three infinite series: $x = \sum_{n=0}^\infty a^n$ $y = \sum_{n=0}^\infty b^n$ $z = \sum_{n=0}^\infty c^n$

Each of these is an infinite geometric progression with the first term as $a^0 = 1$, $b^0 = 1$, and $c^0 = 1$ respectively, and common ratios $a, b, c$ respectively. The problem states that $|a| < 1$, $|b| < 1$, and $|c| < 1$. This ensures that the sums of these infinite G.P.s converge. Using the formula for the sum of an infinite G.P., $S_\infty = \frac{A}{1-R}$, we get: $x = \frac{1}{1-a}$ $y = \frac{1}{1-b}$ $z = \frac{1}{1-c}$ We are also given that $abc \ne 0$. Since $|a|<1, |b|<1, |c|<1$, it implies $a, b, c$ are not equal to 1, so the denominators $1-a, 1-b, 1-c$ are non-zero.

Step 2: Utilizing the A.P. property of a, b, c to form an A.P. of their denominators.

We are given that $a, b, c$ are in A.P. This means that $b-a = c-b$, or $2b = a+c$. We can manipulate this A.P. to get an A.P. involving the denominators of $x, y, z$. If $a, b, c$ are in A.P., then:

1. Multiplying by $-1$: $-a, -b, -c$ are also in A.P. (The common difference becomes $-(b-a)$).
2. Adding $1$: $1-a, 1-b, 1-c$ are also in A.P. (The common difference remains $-(b-a)$).

So, the terms $(1-a), (1-b), (1-c)$ form an arithmetic progression.

Step 3: Relating the A.P. of denominators to the H.P. of x, y, z.

From Step 1, we have $x = \frac{1}{1-a}$, $y = \frac{1}{1-b}$, and $z = \frac{1}{1-c}$. This means that $x, y, z$ are the reciprocals of the terms $(1-a), (1-b), (1-c)$. Since $(1-a), (1-b), (1-c)$ are in A.P. (from Step 2), their reciprocals must be in H.P. Therefore, $x, y, z$ are in Harmonic Progression.

Step 4: Determining the relationship for the reciprocals of x, y, z.

The definition of a Harmonic Progression states that if a sequence is in H.P., then the sequence of its reciprocals is in A.P. Since we have established that $x, y, z$ are in H.P., it directly follows that their reciprocals, $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$, are in A.P.

This corresponds to option (C).

Common Mistakes & Tips

• Confusing H.P. with A.P.: Remember that if terms are in H.P., their reciprocals are in A.P., not the terms themselves.
• Algebraic Manipulation Errors: Carefully apply the properties of A.P. (adding a constant or multiplying by a non-zero constant preserves the A.P. property).
• Ignoring Convergence Conditions: Always ensure that the conditions for the sum of an infinite G.P. ($|R|<1$) are met, as provided in the problem statement.

Summary

The problem involves evaluating infinite geometric series to express $x, y, z$ in terms of $a, b, c$. The given condition that $a, b, c$ are in A.P. is then used to show that $1-a, 1-b, 1-c$ are also in A.P. By the definition of a Harmonic Progression, if the terms $1-a, 1-b, 1-c$ are in A.P., their reciprocals $x, y, z$ must be in H.P. Consequently, the reciprocals of $x, y, z$, which are $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$, must be in A.P.

The final answer is \boxed{C}.