If `a (b +c), b (c+a) , c (a +b)` are in A.P. , prove that `(1)/(a), (1)/(b), (1)/(c )` are also ln A.P.

Answer

To prove that if \( a(b+c), b(c+a), c(a+b) \) are in Arithmetic Progression (A.P.), then \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are also in A.P., we can proceed as follows:

### Step 1: Understand the condition for A.P.
The terms \( a(b+c), b(c+a), c(a+b) \) are in A.P. if:
\[ 2b(c+a) = a(b+c) + c(a+b) \]

### Step 2: Expand the equation
Expanding both sides:
- Left-hand side:
\[ 2b(c+a) = 2bc + 2ab \]
- Right-hand side:
\[ a(b+c) + c(a+b) = ab + ac + ca + cb = ab + ac + bc + ca \]

### Step 3: Set the equation
Setting both sides equal gives:
\[ 2bc + 2ab = ab + ac + bc + ca \]

### Step 4: Rearranging the equation
Rearranging the equation:
\[ 2bc + 2ab - ab - ac - bc - ca = 0 \]
This simplifies to:
\[ bc + ab - ac - ca = 0 \]

### Step 5: Factor the equation
Factoring gives:
\[ ab + bc - ac - ca = 0 \]
This can be rearranged to:
\[ ab + bc = ac + ca \]

### Step 6: Divide by abc
Now, divide the entire equation by \( abc \):
\[ \frac{ab}{abc} + \frac{bc}{abc} = \frac{ac}{abc} + \frac{ca}{abc} \]
This simplifies to:
\[ \frac{1}{c} + \frac{1}{a} = \frac{1}{b} + \frac{1}{b} \]

### Step 7: Rearranging terms
Rearranging gives:
\[ \frac{1}{a} + \frac{1}{c} = 2 \cdot \frac{1}{b} \]

### Step 8: Conclusion
Thus, we have shown that:
\[ 2 \cdot \frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]
This confirms that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P.