If $a + B + C$ Is Not Equal To 0 And $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ Are In Arithmetic Progression, Prove That $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ Are Also In Arithmetic Progression.

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Introduction

In the realm of mathematical problems, questions involving arithmetic progressions (A.P.) often present elegant challenges. This article delves into one such intriguing problem. Given that $a + b + c \neq 0$ and the fractions $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in arithmetic progression, we aim to prove that the reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ also form an arithmetic progression. This exploration will involve leveraging the properties of arithmetic progressions and algebraic manipulations to reach a conclusive proof. Understanding arithmetic progressions is crucial in various mathematical contexts, and this specific problem enhances our ability to work with these sequences in more complex scenarios. Let's embark on this mathematical journey, carefully examining each step to unravel the solution.

Understanding Arithmetic Progressions

Before we tackle the problem directly, it's essential to have a solid understanding of arithmetic progressions (A.P.). An arithmetic progression is a sequence of numbers such that the difference between any two successive members is a constant. This constant difference is called the common difference. For example, the sequence 2, 5, 8, 11, ... is an A.P. with a common difference of 3. A general arithmetic progression can be represented as $a, a+d, a+2d, a+3d, ...$, where $a$ is the first term and $d$ is the common difference. A fundamental property of an A.P. is that for any three consecutive terms, say $x, y, z$, the middle term $y$ is the arithmetic mean of the other two terms. This means that $2y = x + z$. This property is the cornerstone of many A.P. related problems, and we will heavily rely on it in our proof. Understanding this property allows us to manipulate and relate terms within an A.P., which is crucial for solving more complex problems. In the context of our problem, we are given that $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P., and we will use this property to establish a relationship between these terms. Furthermore, knowing the characteristics of arithmetic progressions helps us to discern how transformations of the terms, such as taking reciprocals, might affect the sequence's properties. Thus, a firm grasp of the definition and key properties of arithmetic progressions is vital for solving the given problem.

Setting up the Problem

The problem states that if $a + b + c \neq 0$ and $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in arithmetic progression (A.P.), we need to prove that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are also in A.P. To begin, let's express the given condition mathematically. Since $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P., the middle term is the arithmetic mean of the other two terms. Thus, we can write:

$2(\frac{c+a}{b}) = \frac{b+c}{a} + \frac{a+b}{c}$

This equation forms the basis of our proof. Our goal is to manipulate this equation algebraically to arrive at a condition that shows $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. The condition for $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ to be in A.P. is:

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

Or equivalently:

$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

This is the target equation we need to derive from the given condition. The problem setup involves translating the given information into mathematical equations and identifying the condition we need to prove. This initial setup is crucial because it provides a clear direction for the subsequent algebraic manipulations. In the next sections, we will focus on manipulating the initial equation to ultimately arrive at the target equation, thus proving the required condition. The condition $a + b + c \neq 0$ is also important as it prevents division by zero in certain manipulations.

Algebraic Manipulation

Now, let's focus on the algebraic manipulation of the equation we derived from the given condition:

$2(\frac{c+a}{b}) = \frac{b+c}{a} + \frac{a+b}{c}$

Our aim is to transform this equation into a form that resembles the condition for $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ to be in A.P., which is $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$. To begin, let's multiply both sides of the equation by $abc$ to eliminate the fractions:

$2ac(c+a) = bc(b+c) + ab(a+b)$

Expanding the terms, we get:

$2ac^2 + 2a^2c = b^2c + bc^2 + a^2b + ab^2$

Now, let's rearrange the terms to bring all terms to one side of the equation:

$2ac^2 + 2a^2c - b^2c - bc^2 - a^2b - ab^2 = 0$

To proceed further, we add and subtract $abc$ to facilitate factoring:

$2ac^2 + 2a^2c - b^2c - bc^2 - a^2b - ab^2 - abc + 3abc = 0$

Rearranging the terms strategically:

$(2ac^2 + 2a^2c) - (a^2b + abc + bc^2) - (ab^2 + abc + b^2c) + 3abc = 0$

This step is crucial for identifying common factors and simplifying the expression. Notice that this manipulation involves careful observation and a bit of algebraic intuition. The next step involves factoring and further simplification to reveal the desired relationship. This part of the proof highlights the importance of algebraic proficiency and strategic manipulation to solve mathematical problems. The rearrangements and factoring are key to transforming the initial equation into a more manageable form that will ultimately lead to the solution.

Factoring and Simplification

Continuing from the previous step, we had the equation:

$2ac^2 + 2a^2c - b^2c - bc^2 - a^2b - ab^2 = 0$

After adding and subtracting $3abc$ and strategically rearranging, we aim to factor the expression. Factoring is a key step in simplifying algebraic expressions and revealing underlying relationships. Let's rewrite the equation as:

$2ac(a + c) - [bc(b + c) + ab(a + b)] = 0$

Now, adding and subtracting $abc$ within the brackets to facilitate factoring, we get:

$2ac(a + c) - [bc(b + c) + abc] - [ab(a + b) + abc] + 2abc = 0$

However, a more insightful approach is to directly rearrange the terms and attempt factoring:

$(2a^2c + 2ac^2) - (a^2b + ab^2 + bc^2 + b^2c) = 0$

Rearrange as:

$2ac(a + c) - [a^2b + ab^2 + bc^2 + b^2c] = 0$

This form isn't directly factorable into the terms we need. Let's return to our original expanded form and try a different approach:

$2ac^2 + 2a^2c - b^2c - bc^2 - a^2b - ab^2 = 0$

Add $2abc$ and subtract $2abc$:

$2ac^2 + 2a^2c - b^2c - bc^2 - a^2b - ab^2 + 2abc - 2abc = 0$

Rearrange to attempt factorization:

$(2a^2c - a^2b - abc) + (2ac^2 - ab^2 - abc) - (b^2c + bc^2 - 2abc) = 0$

Factor out common terms:

$a(2ac - ab - bc) + b(2ac - ab - bc) = a^2b + ab^2 + bc^2 + b^2c$

Which doesn't directly lead to a clear factorization. Instead, let's try another rearrangement:

$2ac(a+c) - (b+c)(bc+ab) = 0$

Let's go back to the basics: $2ac^2 + 2a^2c = b^2c + bc^2 + a^2b + ab^2$

Subtract $2abc$ from both sides:

$2ac^2 + 2a^2c + 2abc = b^2c + bc^2 + a^2b + ab^2$

$2ac(a + c) = a^2b + ab^2 + bc^2 + b^2c - 2abc $

This algebraic manipulation highlights a critical aspect of problem-solving: the willingness to revisit and adjust one's approach. Factoring, though a powerful technique, requires careful arrangement and sometimes, alternative strategies are necessary to progress towards the solution.

Reaching the Conclusion

After several attempts at factoring and simplifying the equation, we return to the original equation derived from the A.P. condition:

$2(\frac{c+a}{b}) = \frac{b+c}{a} + \frac{a+b}{c}$

Multiply both sides by $abc$ to clear the fractions:

$2ac(c+a) = bc(b+c) + ab(a+b)$

Expand the terms:

$2ac^2 + 2a^2c = b^2c + bc^2 + a^2b + ab^2$

Now, add $2abc$ to both sides of the equation:

$2ac^2 + 2a^2c + 2abc = b^2c + bc^2 + a^2b + ab^2 + 2abc$

This step is crucial as it allows us to manipulate the equation further towards our desired result. Next, rearrange the terms on both sides to prepare for factoring:

$2ac(a + c + b) = b^2c + bc^2 + a^2b + ab^2 + 2abc$

Notice that we can rewrite the right side by grouping terms strategically:

$2ac(a + b + c) = bc(b + c + a) + ab(a + b + c)$

Now, factor out $(a + b + c)$ from the right side:

$2ac(a + b + c) = (a + b + c)(bc + ab)$

Since $a + b + c \neq 0$, we can divide both sides by $(a + b + c)$:

$2ac = bc + ab$

Now, divide both sides by $abc$ (assuming $a, b, c \neq 0$):

$\frac{2ac}{abc} = \frac{bc}{abc} + \frac{ab}{abc}$

Simplify the fractions:

$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

This is precisely the condition for $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ to be in arithmetic progression. Thus, we have successfully proven that if $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P. and $a + b + c \neq 0$, then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are also in A.P.

Conclusion

In conclusion, we have successfully demonstrated that if $a + b + c \neq 0$ and $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in arithmetic progression, then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are also in arithmetic progression. This proof involved a combination of algebraic manipulation, factoring techniques, and a strategic approach to rearranging terms. The key steps included transforming the initial A.P. condition into an algebraic equation, attempting various factoring strategies, and finally, manipulating the equation to arrive at the desired condition for the reciprocals to be in A.P. This problem underscores the importance of persistence and flexibility in problem-solving, as several approaches were attempted before the successful one was found. Understanding the properties of arithmetic progressions and proficiency in algebraic manipulation are essential tools in tackling such problems. The condition $a + b + c \neq 0$ played a crucial role, allowing us to divide by this expression without concern for division by zero. The logical flow from the initial setup to the final conclusion highlights the beauty and rigor of mathematical proofs. This exercise not only reinforces our understanding of arithmetic progressions but also enhances our problem-solving skills in algebra and mathematical reasoning.