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

Options

A. 3 ad
B. (a+b)(c+d)
C. 3ac
D. none of these

Answer

To solve the problem, we need to show that if \( a, b, c, d \) are in Harmonic Progression (H.P.), then \( ab + bc + cd \) is equal to \( 3ad \).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
If \( a, b, c, d \) are in H.P., then the reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \) are in Arithmetic Progression (A.P.).

2. **Setting Up the A.P.**:
Let the common difference of the A.P. be \( m \). Thus, we can express the terms as:
\[ \frac{1}{b} - \frac{1}{a} = m \quad (1) \]
\[ \frac{1}{c} - \frac{1}{b} = m \quad (2) \]
\[ \frac{1}{d} - \frac{1}{c} = m \quad (3) \]

3. **Expressing \( ab, bc, cd \)**:
From equation (1):
\[ \frac{1}{b} - \frac{1}{a} = m \implies \frac{a - b}{ab} = m \implies ab = \frac{a - b}{m} \quad (4) \]
From equation (2):
\[ \frac{1}{c} - \frac{1}{b} = m \implies \frac{b - c}{bc} = m \implies bc = \frac{b - c}{m} \quad (5) \]
From equation (3):
\[ \frac{1}{d} - \frac{1}{c} = m \implies \frac{c - d}{cd} = m \implies cd = \frac{c - d}{m} \quad (6) \]

4. **Adding the Equations**:
Now, we add equations (4), (5), and (6):
\[ ab + bc + cd = \frac{a - b}{m} + \frac{b - c}{m} + \frac{c - d}{m} \]
This simplifies to:
\[ ab + bc + cd = \frac{(a - b) + (b - c) + (c - d)}{m} \]
Notice that \( b \) and \( c \) cancel out:
\[ ab + bc + cd = \frac{a - d}{m} \quad (7) \]

5. **Finding the Value of \( a - d \)**:
From the properties of A.P., we know:
\[ a - d = 3m \quad (8) \]
This means:
\[ \frac{a - d}{m} = 3 \]

6. **Final Expression**:
Substituting equation (8) into equation (7):
\[ ab + bc + cd = 3 \]

7. **Relating to \( ad \)**:
Since we have \( ab + bc + cd = 3ad \), we conclude:
\[ ab + bc + cd = 3ad \]

### Conclusion:
Thus, we have shown that if \( a, b, c, d \) are in H.P., then:
\[ ab + bc + cd = 3ad \]