If `p,q,r,s in N` and the are four consecutive terms of an A.P., then `p^(th),q^(th),r^(th)ands^(th)` terms of a G.P. are in

Options

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

Answer

To solve the problem, we need to determine the relationship between the Pth, Qth, Rth, and Sth terms of a geometric progression (G.P.) when P, Q, R, and S are four consecutive terms of an arithmetic progression (A.P.).

### Step-by-step Solution:

1. **Understanding the A.P.**:
Let the four consecutive terms of the A.P. be represented as:
- \( p = a \)
- \( q = a + d \)
- \( r = a + 2d \)
- \( s = a + 3d \)
where \( a \) is the first term and \( d \) is the common difference.

2. **Finding the G.P. Terms**:
The Pth, Qth, Rth, and Sth terms of a G.P. can be expressed as:
- \( T_P = A \cdot r^{p-1} \)
- \( T_Q = A \cdot r^{q-1} \)
- \( T_R = A \cdot r^{r-1} \)
- \( T_S = A \cdot r^{s-1} \)
where \( A \) is the first term of the G.P. and \( r \) is the common ratio.

3. **Substituting the A.P. Terms**:
Substitute the values of \( p, q, r, s \) from the A.P. into the G.P. terms:
- \( T_P = A \cdot r^{a - 1} \)
- \( T_Q = A \cdot r^{(a + d) - 1} = A \cdot r^{a + d - 1} \)
- \( T_R = A \cdot r^{(a + 2d) - 1} = A \cdot r^{a + 2d - 1} \)
- \( T_S = A \cdot r^{(a + 3d) - 1} = A \cdot r^{a + 3d - 1} \)

4. **Checking for G.P. Condition**:
To check if these terms form a G.P., we need to see if the ratio of consecutive terms is constant:
- The ratio \( \frac{T_Q}{T_P} = \frac{A \cdot r^{a + d - 1}}{A \cdot r^{a - 1}} = r^d \)
- The ratio \( \frac{T_R}{T_Q} = \frac{A \cdot r^{a + 2d - 1}}{A \cdot r^{a + d - 1}} = r^d \)
- The ratio \( \frac{T_S}{T_R} = \frac{A \cdot r^{a + 3d - 1}}{A \cdot r^{a + 2d - 1}} = r^d \)

Since all ratios are equal to \( r^d \), we conclude that \( T_P, T_Q, T_R, T_S \) are in G.P.

5. **Conclusion**:
Therefore, the Pth, Qth, Rth, and Sth terms of the G.P. are in G.P.

### Final Answer:
The Pth, Qth, Rth, and Sth terms of the G.P. are in G.P.