Question

Statement 1 If \(\mathrm{a}^{2}+2 \mathrm{bc}, \mathrm{b}^{2}+2 \mathrm{ac}, \mathrm{c}^{2}+2 \mathrm{ab}\) are in AP then, \(\mathrm{b}-c_{2} \mathrm{c}-\mathrm{a}, \mathrm{a}-\mathrm{b}\) are in \(\mathrm{HP}\) and Statement 2 If \(a_{1}, a_{2}, \ldots . a_{n}\) are in AP then \(\frac{a_{1}+k}{h}, \frac{a_{2}+k}{h}, \ldots \ldots .\) in AP

Short answer

Question: Prove the following statements: 1. If \(a^2+2bc\), \(b^2+2ac\), and \(c^2+2ab\) are in AP, prove that \(b-c\), \(c-a\), and \(a-b\) are in HP. 2. If \(a_1, a_2, ..., a_n\) are in AP, show that the sequence obtained by adding k to each term and then dividing by h is also in AP. Answer: 1. To prove that \(b-c\), \(c-a\), and \(a-b\) are in HP, we show that their reciprocals, i.e., \(\frac{1}{b-c}\), \(\frac{1}{c-a}\), and \(\frac{1}{a-b}\), are in AP. The given condition for AP is \((b^2+2ac)-(a^2+2bc) = (c^2+2ab)-(b^2+2ac)\), which simplifies to \(a-c=b-c+a-b\). The AP condition for reciprocals is \(\frac{1}{b-c} + \frac{1}{a-b} = 2 \cdot \frac{1}{c-a}\), so the result from Step 7 (Step 10 in solution) proves that \(b-c\), \(c-a\), and \(a-b\) are in HP. 2. The original sequence is in AP with common difference d. The transformed sequence has terms \(\frac{a_1+k}{h}\), \(\frac{a_2+k}{h}\), ..., \(\frac{a_n+k}{h}\). The differences for the transformed sequence are \(\frac{a_2-a_1}{h}\), \(\frac{a_3-a_2}{h}\), ..., \(\frac{a_n-a_{n-1}}{h}\). By replacing the common difference d, we get \(\frac{d}{h}, \frac{d}{h}, ...,\frac{d}{h}\), which verifies the AP condition for the transformed sequence.

Step-by-step solution

Write the given equations in AP

We are given that \(a^2+2bc\), \(b^2+2ac\), \(c^2+2ab\) are in AP.

Apply the condition of AP

In an AP, the difference between any two consecutive terms is constant. So, we have: \((b^2+2ac)-(a^2+2bc) = (c^2+2ab)-(b^2+2ac)\)

Simplify the equation

Simplify the equation obtained in Step 2: \(b^2 - a^2 + 2(ac-bc) = c^2 - b^2 + 2(ab-ac)\)

Rearrange the terms

Rearrange the equation obtained in Step 3 as: \((b^2 - a^2) - (c^2 - b^2) = 2(ab - ac - ac + bc)\)

Apply the difference of squares

Apply the difference of squares formula, \(x^2 - y^2 = (x + y)(x - y)\), and simplify: \((b - a)(b + a) - (b - c)(b + c) = 2b(c-a) + 2c(b-a)\)

Divide both sides by (b-a)

Divide both sides of the equation by \((b-a)\): \((b + a) - (b + c) = -2c + 2b\)

Simplify the equation

Simplify the equation obtained in Step 6: \(a - c = b - c + a - b\)

Write the HP condition

When three numbers \(x, y, z\) are in HP, then their reciprocals are in AP: \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in AP. Here, we want to prove that the numbers \(b - c, c - a, a - b\) are in HP.

Apply the condition of AP to the reciprocals of the numbers

Apply the AP condition to the reciprocals of \(b-c, c-a, a-b\) (using the fact that sum of first and third term is equal to twice the middle term in AP): \(\frac{1}{b-c} + \frac{1}{a-b} = 2 \cdot \frac{1}{c-a}\)

Use the result obtained in Step 7

We know that \(a-c = b-c+a-b\). Now, we replace \(c-a\) with this expression in the equation obtained in Step 9: \(\frac{1}{b-c} + \frac{1}{a-b} = 2 \cdot \frac{1}{a-c}\) This equation proves that the numbers \(b - c, c - a, a - b\) are in HP. Statement 1 is proved. Statement 2:

Write the given equations in AP

We are given that \(a_1, a_2,\dots, a_n\) are in AP.

Write the transformed sequence

We need to prove that the transformed sequence \(\frac{a_1+k}{h}, \frac{a_2+k}{h},\dots,\frac{a_n+k}{h}\) is also in AP.

Find the common difference of the original sequence

Let the common difference of the original sequence be d. Then, \(a_2-a_1 = a_3-a_2 = \dots = a_n-a_{n-1} = d.\)

Compute differences for the transformed sequence

Now, let's compute the differences for the transformed sequence: \(\left(\frac{a_2+k}{h}-\frac{a_1+k}{h}\right),\left(\frac{a_3+k}{h}-\frac{a_2+k}{h}\right), \dots, \left(\frac{a_n+k}{h}-\frac{a_{n-1}+k}{h}\right)\)

Simplify the expression

Simplify the expression obtained in Step 4: \(\frac{a_2-a_1}{h}, \frac{a_3-a_2}{h}, \dots,\frac{a_n-a_{n-1}}{h}\)

Replace the common difference

Replace the common difference, d, in the expression obtained in Step 5: \(\frac{d}{h}, \frac{d}{h}, \dots, \frac{d}{h}\)

Verify the condition of AP

Since the differences in the transformed sequence are all equal to \(\frac{d}{h}\), the sequence \(\frac{a_1+k}{h}, \frac{a_2+k}{h},\dots,\frac{a_n+k}{h}\) is in AP. Statement 2 is proved.

Key concepts

Harmonic Progression

A Harmonic Progression (HP) occurs when the reciprocals of a sequence are in Arithmetic Progression (AP). In other words, a sequence of numbers is considered harmonic if its inverses form an arithmetic sequence. To better understand this concept, consider three sequential numbers in HP:

• If the sequence is \(x, y, z\), then the reciprocal sequence \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) must form an AP.

To check for HP, use the rule:

• The sum of the reciprocal of the first and third terms should be twice the reciprocal of the second term. Mathematically, this is expressed as: \(\frac{1}{x} + \frac{1}{z} = 2 \cdot \frac{1}{y}\).

Understanding HP helps simplify complex problems by transforming them into simpler arithmetic sequences.

Difference of Squares

The Difference of Squares is a powerful algebraic formula used to simplify expressions by representing the difference of two squared terms. It states that \(x^2 - y^2 = (x + y)(x - y)\). This can help break down and simplify expressions:

• Useful in rearranging terms while solving equations.
• Provides a straightforward factorization path, making calculations easier.

Applying this formula effectively requires identifying terms in an expression that fit the squared pattern. For example, with terms like \(b^2 - a^2\), factor them into \((b-a)(b+a)\), a simplification that can reveal deeper structure in equations and aid in solving them logically.

Sequence Transformation

Sequence Transformation is the process of converting a given sequence into another form while retaining specific properties like the common difference. In Statement 2, the sequence transformation performed is: \(\frac{a_1+k}{h}, \frac{a_2+k}{h}, \ldots\). Considerations about this transformation include:

• The addition and division operations ensure the sequence remains arithmetic, as the common difference formulas remain intact both pre and post-transformation.
• Understanding this aids in exploring more complex sequences and changes while preserving mathematical relationships.

Sequence transformations simplify the process of exploring and analyzing sequences, especially in finding patterns or proving theorems.

Common Difference

In arithmetic sequences, the Common Difference is a constant value separating each term from its predecessor and successor. Recognizing this is crucial since it defines the linearity of the sequence:

• This difference \(d\) can be calculated as \(a_2 - a_1 = a_3 - a_2\), and so on, reflecting the sequence's uniformity.
• In transformations, if \(d\) is preserved, the sequence remains arithmetic.

Typically, the common difference is a key parameter in both generating new terms and reversing transformations. Understanding it ensures effective manipulation and accurate conclusions about the nature or properties of arithmetic sequences.