An A.P., and a H.P. have the same first and last terms and the same odd number of terms. The middle terms of the three series are in

Options

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

Answer

To solve the problem, we need to find the middle terms of an Arithmetic Progression (A.P.), a Geometric Progression (G.P.), and a Harmonic Progression (H.P.) that have the same first and last terms and the same odd number of terms. We will then determine the relationship between these middle terms.

### Step-by-Step Solution:

1. **Define the Terms**:
- Let the first term of the A.P. and H.P. be \( a \) and the last term be \( b \).
- The number of terms in each series is \( 2n + 1 \), where \( n \) is a non-negative integer.

2. **Find the Middle Term of the A.P.**:
- The middle term of an A.P. with \( 2n + 1 \) terms is given by:
\[ \text{Middle term of A.P.} = \frac{a + b}{2} \]

3. **Find the Middle Term of the G.P.**:
- The middle term of a G.P. with the same first and last terms is given by:
\[ \text{Middle term of G.P.} = \sqrt{ab} \]

4. **Find the Middle Term of the H.P.**:
- The middle term of an H.P. is the reciprocal of the average of the reciprocals of the first and last terms:
\[ \text{Middle term of H.P.} = \frac{2ab}{a + b} \]

5. **Set Up the Equations**:
- Let:
- \( M_A = \frac{a + b}{2} \) (Middle term of A.P.)
- \( M_G = \sqrt{ab} \) (Middle term of G.P.)
- \( M_H = \frac{2ab}{a + b} \) (Middle term of H.P.)

6. **Relate the Middle Terms**:
- We need to check if the middle terms \( M_A, M_G, M_H \) are in a specific order (A.P., G.P., or H.P.).
- To do this, we will check if \( M_A^2 = M_G \cdot M_H \):
\[ \left(\frac{a + b}{2}\right) \cdot \left(\frac{2ab}{a + b}\right) = ab \]
- Simplifying:
\[ \frac{(a + b) \cdot 2ab}{2(a + b)} = ab \]
- This shows that:
\[ M_A^2 = M_G \cdot M_H \]

7. **Conclusion**:
- Since \( M_A^2 = M_G \cdot M_H \), we conclude that the middle terms \( M_A, M_G, M_H \) are in a G.P.

### Final Answer:
The middle terms of the three series are in a G.P.

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