If x, y, z, w `in` N be four consecutive terms of an A.P., then `T_(x), T_(y), T_(z) and T_(w)` of a G.P. are in

Options

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

Answer

To solve the problem, we need to show that if \( x, y, z, w \) are four consecutive terms of an arithmetic progression (A.P.), then the terms \( T_x, T_y, T_z, T_w \) form a geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding A.P.**:
Since \( x, y, z, w \) are consecutive terms of an A.P., we can express the relationships between these terms:
\[ 2y = x + z \quad \text{(1)} \]
\[ 2z = y + w \quad \text{(2)} \]

2. **Defining G.P. Terms**:
The terms \( T_x, T_y, T_z, T_w \) can be expressed in terms of a first term \( a \) and a common ratio \( r \):
\[ T_x = a r^{x-1}, \quad T_y = a r^{y-1}, \quad T_z = a r^{z-1}, \quad T_w = a r^{w-1} \]

3. **Finding \( T_y^2 \)**:
We calculate \( T_y^2 \):
\[ T_y^2 = (a r^{y-1})^2 = a^2 r^{2y - 2} \]

4. **Substituting \( 2y \)**:
Using equation (1), we substitute \( 2y \):
\[ T_y^2 = a^2 r^{x + z - 2} \]

5. **Rearranging**:
We can express this as:
\[ T_y^2 = a^2 r^{x-1} \cdot r^{z-1} = T_x \cdot T_z \]
This shows that:
\[ T_y^2 = T_x \cdot T_z \quad \text{(3)} \]

6. **Finding \( T_z^2 \)**:
Now, we calculate \( T_z^2 \):
\[ T_z^2 = (a r^{z-1})^2 = a^2 r^{2z - 2} \]

7. **Substituting \( 2z \)**:
Using equation (2), we substitute \( 2z \):
\[ T_z^2 = a^2 r^{y + w - 2} \]

8. **Rearranging Again**:
We can express this as:
\[ T_z^2 = a^2 r^{y-1} \cdot r^{w-1} = T_y \cdot T_w \]
This shows that:
\[ T_z^2 = T_y \cdot T_w \quad \text{(4)} \]

9. **Conclusion**:
From equations (3) and (4), we have:
- \( T_x, T_y, T_z \) are in G.P. because \( T_y^2 = T_x \cdot T_z \).
- \( T_y, T_z, T_w \) are in G.P. because \( T_z^2 = T_y \cdot T_w \).

Thus, we conclude that \( T_x, T_y, T_z, T_w \) are in G.P.