If first and `(2n-1)^th` terms of an AP, GP. and HP. are equal and their nth terms are a, b, c respectively, then (a) a=b=c (b)a+c=b (c) a>b>c and `ac-b^2=0` (d) none of these

Options

A. `a+c = 2b`
B. `a+c=b`
C. `ac-b^(2)=0`
D. none of these

Answer

To solve the problem, we need to analyze the relationships between the first and the \((2n-1)^{th}\) terms of an Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP).

### Step-by-step Solution:

1. **Understanding the Terms**:
- Let the first term of the AP be \(x\) and the \((2n-1)^{th}\) term of the AP be \(y\).
- The first term of the GP is also \(x\) and the \((2n-1)^{th}\) term is \(y\).
- The first term of the HP is also \(x\) and the \((2n-1)^{th}\) term is \(y\).

2. **Setting up the nth Terms**:
- The nth term of the AP is given as \(a\).
- The nth term of the GP is given as \(b\).
- The nth term of the HP is given as \(c\).

3. **Expressing the nth Terms**:
- For the AP, the nth term \(a\) can be expressed as:
\[ a = \frac{x + y}{2} \]
- For the GP, the nth term \(b\) is:
\[ b = \sqrt{xy} \]
- For the HP, the nth term \(c\) is:
\[ c = \frac{2xy}{x + y} \]

4. **Using the Inequalities**:
- By the properties of means, we know that:
\[ a \geq b \geq c \]
- This means:
\[ a \geq b \quad \text{and} \quad b \geq c \]

5. **Using the Relationship Between Means**:
- We also know that the relationship between the means gives us:
\[ b^2 = ac \]
- This is derived from the fact that the square of the geometric mean is equal to the product of the arithmetic mean and the harmonic mean.

6. **Conclusion**:
- From the above analysis, we can conclude that:
- \(a = b = c\) is not necessarily true.
- \(a + c = 2b\) is true.
- \(a \geq b \geq c\) is true.
- \(ac = b^2\) is also true.

### Final Answer:
Thus, the correct options are:
- (b) \(a + c = 2b\)
- (c) \(a \geq b \geq c\) and \(ac = b^2\)