If `T_(p), T_(q), T_(r)` of an A.P. (G.P. or H.P.) are in A.P. (G.P. or H.P.) then p, q, r are in

Options

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

Answer

To solve the problem, we need to analyze the given condition: if \( T_p, T_q, T_r \) of an A.P. (Arithmetic Progression) are in A.P., G.P. (Geometric Progression), or H.P. (Harmonic Progression), then we need to determine the relationship between \( p, q, r \).

### Step-by-Step Solution:

1. **Understanding the Terms**:
- The \( n \)-th term of an A.P. is given by:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Expressing the Terms**:
- For \( T_p, T_q, T_r \):
\[ T_p = a + (p-1)d \]
\[ T_q = a + (q-1)d \]
\[ T_r = a + (r-1)d \]

3. **Condition for A.P.**:
- The terms \( T_p, T_q, T_r \) are in A.P. if:
\[ 2T_q = T_p + T_r \]

4. **Substituting the Expressions**:
- Substitute the expressions for \( T_p, T_q, T_r \):
\[ 2(a + (q-1)d) = (a + (p-1)d) + (a + (r-1)d) \]

5. **Simplifying the Equation**:
- This simplifies to:
\[ 2a + 2(q-1)d = 2a + (p-1)d + (r-1)d \]
- Cancelling \( 2a \) from both sides:
\[ 2(q-1)d = (p-1)d + (r-1)d \]

6. **Factoring Out \( d \)**:
- Assuming \( d \neq 0 \) (since if \( d = 0 \), all terms are equal, which is a trivial case):
\[ 2(q-1) = (p-1) + (r-1) \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ 2q - 2 = p + r - 2 \]
- Thus:
\[ 2q = p + r \]

8. **Conclusion**:
- The equation \( 2q = p + r \) indicates that \( p, q, r \) are in A.P.

### Final Answer:
If \( T_p, T_q, T_r \) of an A.P. are in A.P., then \( p, q, r \) are in A.P.