If a,b,c are in AP and b,c,d be in HP, then

Options

A. `ab=cd`
B. `ad=bc`
C. `ac=bd`
D. `abcd=1`

Answer

To solve the problem step by step, we will use the properties of arithmetic progression (AP) and harmonic progression (HP).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have three terms \( a, b, c \) in AP.
- We also have three terms \( b, c, d \) in HP.

2. **Using the Definition of AP**:
- Since \( a, b, c \) are in AP, we know that:
\[ b = \frac{a + c}{2} \]
- Rearranging gives:
\[ a + c = 2b \quad \text{(Equation 1)} \]

3. **Using the Definition of HP**:
- Since \( b, c, d \) are in HP, we can express this as:
\[ \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \text{ are in AP} \]
- Therefore, we have:
\[ \frac{1}{c} = \frac{\frac{1}{b} + \frac{1}{d}}{2} \]
- This can be rearranged to:
\[ 2 \cdot \frac{1}{c} = \frac{1}{b} + \frac{1}{d} \]
- Multiplying through by \( bcd \) gives:
\[ 2bd = cd + bc \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**:
- From Equation 1, we know \( a + c = 2b \).
- Now, we can multiply both sides of Equation 1 by \( 2b \):
\[ 2b \cdot 2b = (a + c) \cdot 2b \]
- This simplifies to:
\[ 4b^2 = 2ab + 2bc \]

5. **Rearranging Equation 2**:
- From Equation 2, we can rearrange to find:
\[ 2bd - bc = cd \]
- This can be rewritten as:
\[ 2bd = bc + cd \]

6. **Cross Multiplying**:
- We can now multiply the two equations derived:
\[ 2b \cdot c = (a + c) \cdot d \]
- This leads to:
\[ 2bc = ad + cd \]

7. **Final Rearrangement**:
- Rearranging gives:
\[ ad = 2bc - cd \]
- Since \( cd \) cancels out, we arrive at:
\[ ad = bc \]

### Conclusion:
Thus, we have proved that:
\[ AD = BC \]