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 follow these steps:
### Step 1: Understand the condition for A.P.
Given that \( a(b+c), b(c+a), c(a+b) \) are in A.P., we know that:
\[
2b(c+a) = a(b+c) + c(a+b)
\]
### Step 2: Expand the equation
Expanding both sides:
- Left-hand side:
\[
2b(c+a) = 2bc + 2ab
\]
- Right-hand side:
\[
a(b+c) + c(a+b) = ab + ac + ca + cb = ab + ac + bc + ab = 2ab + ac + bc
\]
### Step 3: Set the equation
Now we equate both sides:
\[
2bc + 2ab = 2ab + ac + bc
\]
### Step 4: Simplify the equation
Subtract \( 2ab \) from both sides:
\[
2bc = ac + bc
\]
### Step 5: Rearranging terms
Rearranging gives:
\[
2bc - bc = ac
\]
\[
bc = ac
\]
### Step 6: Divide by abc
Now, we divide the entire equation by \( abc \) (assuming \( a, b, c \neq 0 \)):
\[
\frac{bc}{abc} = \frac{ac}{abc}
\]
This simplifies to:
\[
\frac{1}{a} = \frac{1}{c} + \frac{2}{b}
\]
### Step 7: Rearranging to show A.P.
Rearranging the above equation gives:
\[
\frac{1}{a} + \frac{1}{c} = 2 \cdot \frac{1}{b}
\]
This shows that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P. since it satisfies the condition for A.P. where the sum of the first and last term equals twice the middle term.
### Conclusion
Thus, we have proved that if \( a(b+c), b(c+a), c(a+b) \) are in A.P., then \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are also in A.P.
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