Question

If \(\frac{1}{\mathrm{~b}+\mathrm{c}}, \frac{1}{\mathrm{c}+\mathrm{a}}\) and \(\frac{1}{\mathrm{a}+\mathrm{b}}\) are in A.P, then \(\mathrm{a}^{2}, \mathrm{~b}^{2}\) and \(\mathrm{c}^{2}\) are in (1) Geometric Progression (2) Arithmetic Progression (3) Harmonic Progression (4) None of these

Short answer

Question: If $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in Arithmetic Progression, then $a^2, b^2, c^2$ are in: 1. Geometric Progression 2. Arithmetic Progression 3. Harmonic Progression 4. None of these Answer: (4) None of these

Step-by-step solution

Use the given information

We are given that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in AP. So, we have: $$\frac{1}{c+a} - \frac{1}{b+c} = \frac{1}{a+b} - \frac{1}{c+a}$$

Simplify the equation

Now we will simplify the equation: $$\frac{(b+c)-(c+a)}{(b+c)(c+a)} = \frac{(a+b)-(c+a)}{(a+b)(c+a)}$$

Simplify further and solve for a, b, and c

Now simplify further: $$\frac{b-a}{(a+b)(b+c)} = \frac{a-c}{(a+b)(c+a)}$$ Cross-multiply and we get: $$(a-c)(a+b)(b+c) = (b-a)(a+b)(c+a)$$ Now, notice that if we add \(b + c\) to both sides, they will cancel out: $$b^2(a-c) = a^2(c-a)$$ Since \(a\) and \(c\) are not equal, we can safely divide both sides by \((a-c)\) and we get: $$b^2 = ac$$

Test if a², b², c² are in Geometric Progression

To test if \(a^2, b^2\), and \(c^2\) are in GP, we will see if the ratio of \(b^2\) to \(a^2\) is equal to the ratio of \(c^2\) to \(b^2\): $$\frac{b^2}{a^2} = \frac{c^2}{b^2}$$ Using the result from Step 3: $$\frac{ac}{a^2} = \frac{c^2}{ac}$$ Upon further simplifying, we get: $$\frac{1}{a} = \frac{c}{a}$$ Which is not true, so \(a^2, b^2, c^2\) are not in Geometric Progression.

Test if a², b², c² are in Arithmetic Progression

To test if \(a^2, b^2\), and \(c^2\) are in AP, we can check if: $$b^2 - a^2 = c^2 - b^2$$ Using the result from Step 3, \(b^2 = ac\), we can plug it into the equation: $$ac - a^2 = c^2 - ac$$ Which simplifies to: $$2ac - a^2 = c^2$$ This equation does not hold for all values of \(a\) and \(c\), so \(a^2, b^2, c^2\) are not in Arithmetic Progression.

Test if a², b², c² are in Harmonic Progression

To test if \(a^2, b^2\), and \(c^2\) are in Harmonic Progression (HP), we will see if their reciprocals are in AP: $$\frac{1}{a^2}, \frac{1}{b^2}, \frac{1}{c^2}$$ We already know that \(\frac{1}{b^2} = \frac{1}{ac}\), so we can plug it in: $$\frac{1}{a^2}, \frac{1}{ac}, \frac{1}{c^2}$$ It can be seen that they are not in AP, so \(a^2, b^2, c^2\) are not in Harmonic Progression.

Conclusion

Since we have shown that \(a^2, b^2, c^2\) are not in Geometric Progression, Arithmetic Progression, or Harmonic Progression, the answer is: (4) None of these

Key concepts

Geometric Progression

Geometric Progression, often abbreviated as GP, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This concept is fundamental in mathematics and appears frequently in mathematical problem solving and real-life applications. In GP, if you start with the first term \(a\) and the common ratio \(r\), the sequence is \(a, ar, ar^2, ar^3, \ldots\). Each term is derived from the prior term by multiplying it with \(r\).

**How to Identify a GP:**

• Calculate the ratio of consecutive terms in the sequence.
• If the ratio is the same throughout, it's a geometric progression.

In the earlier exercise, we tested whether the squares of three variables, \(a^2, b^2, ext{ and } c^2\), were in GP by checking if the ratios of successive terms were equal. We concluded they were not in GP because the condition \(\frac{b^2}{a^2} = \frac{c^2}{b^2}\) did not hold.

Harmonic Progression

Harmonic Progression, or HP, is less common yet an interesting type of sequence. In an HP, the reciprocals of the terms form an arithmetic progression (AP). This means that if the sequence is \(a, b, c, \ldots\), then \(\frac{1}{a}, \frac{1}{b},\frac{1}{c}, \ldots\) should be in AP.

**Understanding HP:**

• To check if a sequence forms an HP, verify if the reciprocals form an AP.
• HPs are not particularly common but appear in specific mathematical scenarios like solving certain types of equations.

In the exercise, we looked to see if \(a^2, b^2, c^2\) were in HP by checking if \(\frac{1}{a^2}, \frac{1}{b^2}, \frac{1}{c^2}\) were in AP. They were not, indicating that \(a^2, b^2, c^2\) does not form an HP.

Mathematical Problem Solving

Mathematical Problem Solving is a crucial skill that involves a process to identify the best methods or operations needed to solve a given problem. The approach depends on understanding the problem, devising a plan, carrying out the plan, and evaluating the solution.

**Steps for Effective Problem Solving:**

• Understand the Problem: Carefully read and identify what is given and what needs to be determined.
• Devise a Plan: Think of possible strategies, like simplifying the problem, working backward, or drawing diagrams.
• Execute the Plan: Follow through with the methods chosen. This step involves calculations and manipulations.
• Reflect: Check if the solution makes sense and is consistent with the problem statement. If required, revise the approach.

The exercise provided a platform to employ these problem-solving strategies starting from interpreting the conditions for AP and exploring different progressions for \(a^2, b^2, c^2\), leading to a thoughtful conclusion.