Math Problem Statement

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.

Solution

We are given that the terms:

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

are in arithmetic progression (A.P.). To prove that a2a^2, b2b^2, and c2c^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:

21c+a=1b+c+1a+b2 \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:

21c+a=1b+c+1a+b2 \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:

1b+c+1a+b=(a+b)+(b+c)(b+c)(a+b)=a+b+b+c(b+c)(a+b)=a+2b+c(b+c)(a+b)\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:

21c+a=a+2b+c(b+c)(a+b)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)2(c + a)(b + c)(a + b) = (c + a)(a + 2b + c)

Canceling out the common term (c+a)(c + a) on both sides (assuming c+a0c + a \neq 0):

2(b+c)(a+b)=a+2b+c2(b + c)(a + b) = a + 2b + c

Step 4: Simplify

Expand both sides:

2(b+c)(a+b)=2(ab+b2+ac+bc)2(b + c)(a + b) = 2(ab + b^2 + ac + bc) a+2b+c=a+2b+ca + 2b + c = a + 2b + c

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

Step 5: Conclusion for a2,b2,c2a^2, b^2, c^2

Since we have shown that the given terms are in arithmetic progression, we now claim that a2a^2, b2b^2, and c2c^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, a2,b2,c2a^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=ca = 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,ca, b, c were zero?
  5. How would the proof change if we considered negative values for a,b,ca, b, c?

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

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Algebra
Reciprocal Functions

Formulas

Arithmetic progression condition: 2 * second term = first term + third term
Reciprocal sum: 1/(b + c) + 1/(a + b) = (a + 2b + c) / ((b + c)(a + b))

Theorems

If the reciprocals of three numbers are in arithmetic progression, then the squares of those numbers are also in arithmetic progression

Suitable Grade Level

Grade 11-12