If first and (2n−1)^(th) terms of an A.P., G.P. and H.P. are equal and their n^(th) terms are a, b, c respectively, then

Options

A. a. a+c=2b
B. b. a+c=b
C. c. b^2 =ac
D. d. none of these

Answer

To solve the problem, we need to analyze the conditions given for the A.P. (Arithmetic Progression), G.P. (Geometric Progression), and H.P. (Harmonic Progression).

### Step-by-Step Solution:

1. **Define the Terms of A.P., G.P., and H.P.:**
- Let the first term of the A.P. be \( a_1 = x \) and the second term be \( a_2 = y \).
- The \( n^{th} \) term of the A.P. is given by:
\[ a_n = \frac{x + y}{2} \]

2. **For the G.P.:**
- The first term of the G.P. is also \( b_1 = x \) and the second term is \( b_2 = y \).
- The \( n^{th} \) term of the G.P. is given by:
\[ b_n = \sqrt{xy} \]

3. **For the H.P.:**
- The first term of the H.P. is \( c_1 = x \) and the second term is \( c_2 = y \).
- The \( n^{th} \) term of the H.P. is given by:
\[ c_n = \frac{2xy}{x + y} \]

4. **Equating Terms:**
- According to the problem, the first term and the \( (2n-1)^{th} \) term of A.P., G.P., and H.P. are equal. Therefore:
\[ x = a_{2n-1} = \frac{x + (x + (2n-2)d)}{2} \]
\[ x = b_{2n-1} = x \cdot r^{2n-2} \]
\[ x = c_{2n-1} = \frac{2xy}{x + y} \]

5. **Using the Relationships:**
- From the properties of A.M., G.M., and H.M., we know:
\[ a_n \geq b_n \geq c_n \]
- This implies:
\[ \frac{x + y}{2} \geq \sqrt{xy} \geq \frac{2xy}{x + y} \]

6. **Squaring the G.M. Inequality:**
- From the inequality \( \sqrt{xy} \leq \frac{x + y}{2} \), squaring both sides gives:
\[ xy \leq \frac{(x + y)^2}{4} \]
- Rearranging gives:
\[ 4xy \leq x^2 + 2xy + y^2 \implies 0 \leq x^2 - 2xy + y^2 \implies 0 \leq (x - y)^2 \]
- This shows that \( x = y \).

7. **Conclusion:**
- Since \( x = y \), we can substitute back into our equations for \( a_n, b_n, c_n \):
\[ a_n = b_n = c_n \]
- Therefore, the relationship \( b^2 = ac \) holds true.

### Final Answer:
The correct relationship is \( b^2 = ac \).