If `x = underset(n-0)overset(oo)sum a^(n), y= underset(n =0)overset(oo)sum b^(n), z = underset(n =0)overset(oo)sum C^(n)` where a,b,c are in A.P. and `|a| lt 1, |b| lt 1, |c| lt 1`, then x,y,z are in

Options

A. A.P.
B. G.P
C. H.P.
D. none of these

Answer

To solve the problem, we need to analyze the sums \( x \), \( y \), and \( z \) given the conditions that \( a \), \( b \), and \( c \) are in arithmetic progression (A.P.) and that \( |a| < 1 \), \( |b| < 1 \), and \( |c| < 1 \).

### Step-by-Step Solution:

1. **Identify the Series:**
- We have:
\[ x = \sum_{n=0}^{\infty} a^n, \quad y = \sum_{n=0}^{\infty} b^n, \quad z = \sum_{n=0}^{\infty} c^n \]
- Each of these series is a geometric series.

2. **Calculate \( x \):**
- The sum of an infinite geometric series is given by:
\[ S = \frac{a_1}{1 - r} \]
where \( a_1 \) is the first term and \( r \) is the common ratio.
- For \( x \):
\[ x = \frac{1}{1 - a} \quad \text{(since \( a_1 = 1 \) and \( r = a \))} \]

3. **Calculate \( y \):**
- Similarly, for \( y \):
\[ y = \frac{1}{1 - b} \quad \text{(since \( a_1 = 1 \) and \( r = b \))} \]

4. **Calculate \( z \):**
- For \( z \):
\[ z = \frac{1}{1 - c} \quad \text{(since \( a_1 = 1 \) and \( r = c \))} \]

5. **Use the A.P. Condition:**
- Since \( a, b, c \) are in A.P., we have:
\[ 2b = a + c \]
- Rearranging gives:
\[ a + c = 2b \]

6. **Manipulate the Expressions:**
- We can express \( 1 - a \), \( 1 - b \), and \( 1 - c \):
\[ 1 - a = 1 - b + (b - a) \quad \text{and} \quad 1 - c = 1 - b + (b - c) \]
- From the A.P. condition:
\[ 1 - a + 1 - c = 2(1 - b) \]

7. **Show \( x, y, z \) are in Harmonic Progression (H.P.):**
- The terms \( \frac{1}{1 - a}, \frac{1}{1 - b}, \frac{1}{1 - c} \) will be in H.P. if:
\[ 2 \cdot \frac{1}{1 - b} = \frac{1}{1 - a} + \frac{1}{1 - c} \]
- This follows from the rearrangement of the A.P. condition.

### Conclusion:
Thus, \( x, y, z \) are in Harmonic Progression (H.P.).