If x, y, z are in A.P. then xth, yth and zth terms of any G.P. are in

Options

A. A.P.
B. G.P.
C. reciprocals of these terms are in A.P.
D. none of these

Answer

To solve the problem, we need to show that if \( x, y, z \) are in Arithmetic Progression (A.P.), then the \( x \)-th, \( y \)-th, and \( z \)-th terms of any Geometric Progression (G.P.) are also in G.P.

### Step-by-Step Solution:

1. **Understanding A.P.**:
- Since \( x, y, z \) are in A.P., we can express this relationship mathematically:
\[ 2y = x + z \]

2. **Terms of G.P.**:
- Let the first term of the G.P. be \( a \) and the common ratio be \( r \).
- The \( x \)-th term of the G.P. is given by:
\[ T_x = a r^{x-1} \]
- The \( y \)-th term of the G.P. is:
\[ T_y = a r^{y-1} \]
- The \( z \)-th term of the G.P. is:
\[ T_z = a r^{z-1} \]

3. **Relating the Terms**:
- We need to show that \( T_x, T_y, T_z \) are in G.P. This means we need to prove:
\[ T_y^2 = T_x \cdot T_z \]

4. **Substituting the Terms**:
- Substitute the expressions for \( T_x, T_y, T_z \):
\[ T_y^2 = (a r^{y-1})^2 = a^2 r^{2(y-1)} \]
- For \( T_x \cdot T_z \):
\[ T_x \cdot T_z = (a r^{x-1})(a r^{z-1}) = a^2 r^{(x-1) + (z-1)} = a^2 r^{x + z - 2} \]

5. **Using the A.P. Condition**:
- From the A.P. condition \( 2y = x + z \), we can express \( x + z \) as:
\[ x + z = 2y \]
- Substitute this into the equation for \( T_x \cdot T_z \):
\[ T_x \cdot T_z = a^2 r^{2y - 2} \]

6. **Final Comparison**:
- Now we have:
\[ T_y^2 = a^2 r^{2(y-1)} = a^2 r^{2y - 2} \]
- And:
\[ T_x \cdot T_z = a^2 r^{2y - 2} \]
- Since \( T_y^2 = T_x \cdot T_z \), we conclude that \( T_x, T_y, T_z \) are in G.P.

### Conclusion:
Thus, if \( x, y, z \) are in A.P., then the \( x \)-th, \( y \)-th, and \( z \)-th terms of any G.P. are also in G.P.