If for an exponential function `y = a^(x) (a gt 0, ne 1) x_(1), x_(2),…x_(n)` form an A.P., then `y_(1), y_(2),…y_(n)` form a

Options

A. A.P.
B. G.P.
C. H.P.
D. A.G.P.

Answer

To solve the problem, we need to show that if \( x_1, x_2, \ldots, x_n \) are in an arithmetic progression (A.P.), then the corresponding values \( y_1, y_2, \ldots, y_n \) defined by the exponential function \( y = a^x \) will form a geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
Since \( x_1, x_2, \ldots, x_n \) are in A.P., we can express this condition mathematically. For any three terms \( x_1, x_2, x_3 \) in A.P., we have:
\[ 2x_2 = x_1 + x_3 \]

2. **Expressing \( y \) in terms of \( x \)**:
The corresponding values of \( y \) are given by:
\[ y_1 = a^{x_1}, \quad y_2 = a^{x_2}, \quad y_3 = a^{x_3} \]

3. **Using the A.P. Condition**:
From the A.P. condition, we can manipulate the expression for \( y_2 \):
\[ y_2^2 = (a^{x_2})^2 = a^{2x_2} \]
Using the A.P. condition \( 2x_2 = x_1 + x_3 \), we can rewrite this as:
\[ y_2^2 = a^{x_1 + x_3} = a^{x_1} \cdot a^{x_3} = y_1 \cdot y_3 \]

4. **Conclusion for Three Terms**:
The equation \( y_2^2 = y_1 \cdot y_3 \) shows that \( y_1, y_2, y_3 \) are in G.P.

5. **Generalizing to n Terms**:
By applying the same reasoning to all \( n \) terms, we can conclude that if \( x_1, x_2, \ldots, x_n \) are in A.P., then \( y_1, y_2, \ldots, y_n \) will form a G.P.

### Final Result:
Thus, we can conclude that if \( x_1, x_2, \ldots, x_n \) form an A.P., then \( y_1, y_2, \ldots, y_n \) form a G.P.