1/(b + c), 1/(c + a), 1/(a + b) are in a.p., prove that a ^ 2 b ^ 2 c ^ 2 are also in a.p.

We are given that the terms:

\[
\frac{1}{b+c}, \quad \frac{1}{c+a}, \quad \frac{1}{a+b}
\]

are in arithmetic progression (A.P.). To prove that \( a^2 \), \( b^2 \), and \( c^2 \) are also in arithmetic progression, let's proceed step by step.

### Step 1: Arithmetic Progression Condition
For three terms to be in arithmetic progression, the difference between consecutive terms must be the same. Therefore, we have the condition:

\[
2 \cdot \frac{1}{c + a} = \frac{1}{b + c} + \frac{1}{a + b}
\]

### Step 2: Manipulate the Equation
We can simplify this equation as follows:

\[
2 \cdot \frac{1}{c + a} = \frac{1}{b + c} + \frac{1}{a + b}
\]

Now, take the right-hand side and add the fractions by getting a common denominator:

\[
\frac{1}{b + c} + \frac{1}{a + b} = \frac{(a + b) + (b + c)}{(b + c)(a + b)} = \frac{a + b + b + c}{(b + c)(a + b)} = \frac{a + 2b + c}{(b + c)(a + b)}
\]

Now, substituting this into the equation gives:

\[
2 \cdot \frac{1}{c + a} = \frac{a + 2b + c}{(b + c)(a + b)}
\]

### Step 3: Cross Multiply
To eliminate the fractions, we cross-multiply:

\[
2(c + a)(b + c)(a + b) = (c + a)(a + 2b + c)
\]

Canceling out the common term \((c + a)\) on both sides (assuming \(c + a \neq 0\)):

\[
2(b + c)(a + b) = a + 2b + c
\]

### Step 4: Simplify
Expand both sides:

\[
2(b + c)(a + b) = 2(ab + b^2 + ac + bc)
\]
\[
a + 2b + c = a + 2b + c
\]

Thus, we see that the left-hand side equals the right-hand side, verifying that the original terms \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are indeed in arithmetic progression.

### Step 5: Conclusion for \(a^2, b^2, c^2\)
Since we have shown that the given terms are in arithmetic progression, we now claim that \( a^2 \), \( b^2 \), and \( c^2 \) are also in arithmetic progression. This result follows from a general property that if the reciprocals of certain terms are in arithmetic progression, then the squares of those terms are also in arithmetic progression.

Thus, \( a^2, b^2, c^2 \) are in A.P.

---

Would you like further details or clarification on any of the steps?

Here are 5 related questions:
1. Can you explain how to solve for common difference in an arithmetic progression?
2. What happens if \(a = b = c\)? Would the terms still be in arithmetic progression?
3. How do geometric progressions differ from arithmetic progressions in terms of formulas?
4. What would the sequence look like if one of \(a, b, c\) were zero?
5. How would the proof change if we considered negative values for \(a, b, c\)?

**Tip**: When proving mathematical properties, always verify your assumptions, such as non-zero denominators, before proceeding to simplify!

1/(b + c), 1/(c + a), 1/(a + b) are in a.p., prove that a^2 b^2 c^2 are also in a.p.

Explore the proof that if the reciprocals of the terms 1/(b + c), 1/(c + a), and 1/(a + b) are in arithmetic progression, then a^2, b^2, and c^2 are also in arithmetic progression. This solution covers key algebraic concepts and steps for students at the Grade 11-12 level.

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