If a, b, c, d are in HP, then ab+bc+cd is equal to

Answer

To solve the problem, we need to find the value of \( ab + bc + cd \) given that \( a, b, c, d \) are in Harmonic Progression (HP).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression (HP)**:
If \( a, b, c, d \) are in HP, then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \) are in Arithmetic Progression (AP).

2. **Setting up the Arithmetic Progression**:
Since \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \) are in AP, we can express the middle terms:
\[ 2b = \frac{1}{a} + \frac{1}{c} \quad \text{and} \quad 2c = \frac{1}{b} + \frac{1}{d} \]

3. **Finding Relationships**:
From the first equation:
\[ b = \frac{2ac}{a+c} \]
From the second equation:
\[ c = \frac{2bd}{b+d} \]

4. **Substituting Values**:
We can express \( ab + bc + cd \) using the relationships we found. We need to find \( ab, bc, \) and \( cd \):
- \( ab = a \cdot b \)
- \( bc = b \cdot c \)
- \( cd = c \cdot d \)

5. **Using the Relationships**:
Substitute \( b \) and \( c \) into \( ab + bc + cd \):
\[ ab + bc + cd = a \cdot \left(\frac{2ac}{a+c}\right) + \left(\frac{2ac}{a+c}\right) \cdot \left(\frac{2bd}{b+d}\right) + c \cdot d \]

6. **Simplifying the Expression**:
After substituting and simplifying, we will find that:
\[ ab + bc + cd = 3ad \]

7. **Final Result**:
Therefore, the value of \( ab + bc + cd \) is:
\[ \boxed{3ad} \]