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 find the value of \( ab + bc + cd \) given that \( a, b, c, d \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding H.P.**:
If \( a, b, c, d \) are in H.P., then their 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.**:
Since \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} \) are in A.P., we can express the common difference \( D \):
\[ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{c} = D \]

3. **Finding expressions for \( ab, bc, cd \)**:
- From the first two terms:
\[ \frac{1}{b} - \frac{1}{a} = D \implies \frac{a - b}{ab} = D \implies ab = \frac{a - b}{D} \]
- From the second and third terms:
\[ \frac{1}{c} - \frac{1}{b} = D \implies \frac{b - c}{bc} = D \implies bc = \frac{b - c}{D} \]
- From the third and fourth terms:
\[ \frac{1}{d} - \frac{1}{c} = D \implies \frac{c - d}{cd} = D \implies cd = \frac{c - d}{D} \]

4. **Adding the expressions**:
Now, we add the three expressions:
\[ ab + bc + cd = \frac{a - b}{D} + \frac{b - c}{D} + \frac{c - d}{D} \]
Combining these:
\[ ab + bc + cd = \frac{(a - b) + (b - c) + (c - d)}{D} \]
The terms \( -b \), \( b \), \( -c \), \( c \), \( -d \) cancel out:
\[ ab + bc + cd = \frac{a - d}{D} \]

5. **Relating \( D \) to \( a, d \)**:
From our earlier relationship involving \( D \):
\[ D = \frac{1}{b} - \frac{1}{a} = \frac{a - b}{ab} \]
We can express \( D \) in terms of \( a \) and \( d \) as well.

6. **Final Expression**:
After substituting and simplifying, we find that:
\[ ab + bc + cd = 3AD \]
Thus, the final answer is:
\[ ab + bc + cd = 3AD \]

### Conclusion:
The value of \( ab + bc + cd \) is \( 3AD \).