If `x = sum_(n=0)^(oo) a^(n), y=sum_(n=0)^(oo) b^(n), z = sum_(n=0)^(oo) 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. HP
B. Arithmetic -Geometric Progression
C. AP
D. GP

Answer

To solve the problem, we need to find the sums of the infinite geometric series for \( x \), \( y \), and \( z \), and then determine the relationship between these sums given that \( a, b, c \) are in arithmetic progression (A.P.).

### Step 1: Calculate \( x \)
The sum \( x \) is given by:
\[ x = \sum_{n=0}^{\infty} a^n \]
This is a geometric series with the first term \( 1 \) (when \( n=0 \)) and common ratio \( a \). The formula for the sum of an infinite geometric series is:
\[ S = \frac{a}{1 - r} \]
where \( |r| < 1 \). Here, \( r = a \), so:
\[ x = \frac{1}{1 - a} \]

### Step 2: Calculate \( y \)
Similarly, for \( y \):
\[ y = \sum_{n=0}^{\infty} b^n \]
This is also a geometric series with first term \( 1 \) and common ratio \( b \):
\[ y = \frac{1}{1 - b} \]

### Step 3: Calculate \( z \)
For \( z \):
\[ z = \sum_{n=0}^{\infty} c^n \]
This follows the same pattern:
\[ z = \frac{1}{1 - c} \]

### Step 4: Establish the relationship between \( a, b, c \)
Since \( a, b, c \) are in A.P., we have:
\[ 2b = a + c \]

### Step 5: Express \( x, y, z \) in terms of \( a, b, c \)
From the previous calculations:
\[ x = \frac{1}{1 - a}, \quad y = \frac{1}{1 - b}, \quad z = \frac{1}{1 - c} \]

### Step 6: Show that \( x, y, z \) are in Harmonic Progression (H.P.)
To show that \( x, y, z \) are in H.P., we need to show that:
\[ \frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y} \]

Calculating \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \):
\[ \frac{1}{x} = 1 - a, \quad \frac{1}{y} = 1 - b, \quad \frac{1}{z} = 1 - c \]

Now, substituting these into the H.P. condition:
\[ (1 - b) - (1 - a) = (1 - c) - (1 - b) \]
This simplifies to:
\[ -b + a = -c + b \]
Rearranging gives:
\[ a + c = 2b \]
This confirms that \( x, y, z \) are in H.P.

### Conclusion
Thus, we conclude that if \( a, b, c \) are in A.P., then \( x, y, z \) are in H.P.