Let a,b,c be in A.P. and `|a|lt1,|b|lt1|c|lt1.ifx=1+a+a^(2)+ . . . ."to "oo,y=1+b+b^(2)+ . . . ."to "ooand,z=1+c+c^(2)+ . . . "to "oo`, then x,y,z are in

Options

A. AP
B. GP
C. HP
D. None of these

Explanation

To solve the problem step by step, we will analyze the given information and apply the properties of sequences and series.

### Step 1: Understanding the Given Information
We have three numbers \( a, b, c \) in Arithmetic Progression (A.P.) and the conditions \( |a| < 1 \), \( |b| < 1 \), and \( |c| < 1 \). We also have the following series defined:
- \( x = 1 + a + a^2 + a^3 + \ldots \)
- \( y = 1 + b + b^2 + b^3 + \ldots \)
- \( z = 1 + c + c^2 + c^3 + \ldots \)

### Step 2: Finding the Sum of the Infinite Geometric Series
The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. In our case, the first term is 1, and the common ratio is \( a \), \( b \), or \( c \) respectively.

Thus, we can express \( x, y, z \) as:
\[ x = \frac{1}{1 - a} \]
\[ y = \frac{1}{1 - b} \]
\[ z = \frac{1}{1 - c} \]

### Step 3: Using the Property of A.P.
Since \( a, b, c \) are in A.P., we can express \( b \) and \( c \) in terms of \( a \):
\[ b = a + d \quad \text{and} \quad c = a + 2d \]
for some common difference \( d \).

### Step 4: Finding \( 1 - a, 1 - b, 1 - c \)
Now we can express \( 1 - b \) and \( 1 - c \):
\[ 1 - b = 1 - (a + d) = (1 - a) - d \]
\[ 1 - c = 1 - (a + 2d) = (1 - a) - 2d \]

### Step 5: Establishing the Relationship
Since \( a, b, c \) are in A.P., \( 1 - a, 1 - b, 1 - c \) will also be in A.P. This means that the reciprocals \( \frac{1}{1 - a}, \frac{1}{1 - b}, \frac{1}{1 - c} \) are in Harmonic Progression (H.P.).

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

### Final Answer
Therefore, \( x, y, z \) are in **Harmonic Progression**.

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