Question

Statement 1 If \(b+c, c+a\) and \(a+b\) are in HP then \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) are in HP and Statement 2 If each term of a sequence in a GP is squared, the resulting series is a GP

Short answer

Question: Prove that if \(b+c\), \(c+a\), and \(a+b\) are in Harmonic Progression, then \(\frac{b+c}{a}\), \(\frac{c+a}{b}\), and \(\frac{a+b}{c}\) are also in Harmonic Progression. Also, prove that if each term of a series is in Geometric Progression, the resulting series formed by squaring each term is a Geometric Progression. Answer: We proved that if \(b+c\), \(c+a\), and \(a+b\) are in Harmonic Progression (HP), then \(\frac{b+c}{a}\), \(\frac{c+a}{b}\), and \(\frac{a+b}{c}\) are also in HP by showing that their reciprocals, \(\frac{a}{b+c}\), \(\frac{b}{c+a}\), and \(\frac{c}{a+b}\), are in Arithmetic Progression and have the same common difference as the reciprocals of the given terms. Moreover, we demonstrated that squaring each term of a series in Geometric Progression (GP) results in a series that is also a GP by squaring the general form of the GP, \(ar^{n-1}\), which yields \(a^2r^{2(n-1)}\). This series can be written as a GP with the first term \(a^2\) and common ratio \(r^2\).

Step-by-step solution

Creating expressions for the given HP

We are given that \(b+c, c+a,\) and \(a+b\) are in HP. To work with HP, we can convert them to Arithmetic Progression (AP) by taking their reciprocals. Therefore, sequences \(\frac{1}{b+c}, \frac{1}{c+a},\) and \(\frac{1}{a+b}\) will be in AP.

Finding a common difference for the AP

For any three terms in an AP with a common difference \(d\), we can write the following expressions: \(\frac{1}{c+a} - \frac{1}{b+c} = d\) and \(\frac{1}{a+b} - \frac{1}{c+a} = d\)

Solving for the common difference in terms of \(a, b,\) and \(c\)

To solve for the common difference \(d\), we will find the difference between consecutive terms and simplify the resulting expressions: \(d = \frac{1}{c+a} - \frac{1}{b+c} = \frac{(b+c)-(c+a)}{(b+c)(c+a)} = \frac{b-a}{(b+c)(c+a)}\) Similarly, for the other two terms: \(d = \frac{1}{a+b} - \frac{1}{c+a} = \frac{(c+a)-(a+b)}{(a+b)(c+a)} = \frac{c-b}{(a+b)(c+a)}\)

Applying the condition of the given HP to the desired HP

We want to show that the sequences \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) are also in HP. Following a similar approach as in step 1, we convert this HP to AP using the reciprocals: \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\).

Showing that the AP formed in step 4 has a common difference of \(d\)

For the AP formed in step 4, we need to show that the common difference for these terms is the same \(d\) as derived from the given HP. Finding the difference between the consecutive terms: \(\frac{b}{c+a} - \frac{a}{b+c} = \frac{b^2 - a^2}{(b+c)(c+a)} = \frac{(b-a)(b+a)}{(b+c)(c+a)}\) Similarly, for the other two terms: \(\frac{c}{a+b} - \frac{b}{c+a} = \frac{c^2 - b^2}{(a+b)(c+a)} = \frac{(c-b)(c+b)}{(a+b)(c+a)}\)

Comparing the common differences

Comparing the common differences from the given HP and the desired HP, we can observe the following: 1. \(\frac{b-a}{(b+c)(c+a)} = \frac{(b-a)(b+a)}{(b+c)(c+a)}\) 2. \(\frac{c-b}{(a+b)(c+a)} = \frac{(c-b)(c+b)}{(a+b)(c+a)}\) Dividing the first expression by the second, we get: \(\frac{\frac{b-a}{(b+c)(c+a)}}{\frac{c-b}{(a+b)(c+a)}} = \frac{(b-a)(c+b)}{(c-b)(b+a)}\) Since both sides of the equation are equal, it indicates that both sets of sequences indeed have the same common difference \(d\). If they have the same \(d\), then the sequences \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) must be in AP as well. Therefore, if \(b+c, c+a,\) and \(a+b\) are in HP, then \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\) are also in HP. #Statement 2: Squaring each term of a GP resulted series is a GP#

Defining general forms of a GP

Let the given series in GP be \(ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio.

Squaring each term of the given GP

If we square each term of the given GP, we will get a new series: \((ar^{n-1})^2 = a^2r^{2(n-1)}\)

Proving that the resulting series is a GP

The squared series can be written as a GP with the first term \(a^2\) and the common ratio \(r^2\). So, we can conclude that if each term of a series in a GP is squared, the resulting series is a GP.

Key concepts

Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, known as the common difference, to the previous term. For the sequence to be an AP, this common difference must be the same between any two consecutive terms. For example, the sequence 2, 4, 6, 8 is an arithmetic progression with a common difference of 2.

In the context of the provided exercise, when dealing with a Harmonic Progression (HP), it's beneficial to find the associated AP by taking the reciprocals of the HP terms. Reciprocals of a harmonic sequence form an AP, and this allows us to utilize familiar arithmetic progression formulas and properties to solve problems more easily. This reciprocal relationship is a critical concept since it connects the understanding of HP to the more widely used AP, making complicated HP problems more approachable.

Geometric Progression (GP)

A Geometric Progression (GP) is another type of sequence, but in this case, each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. A simple example of a GP is the sequence 3, 6, 12, 24, where each term is twice the previous term, making the common ratio equal to 2.

In the second statement of the exercise, squaring each term of a GP results in a new GP because the operation of squaring is consistent for every term and hence preserves the multiplicative relationship. Understanding the properties of GPs is essential for recognizing how operations like squaring affect the sequence's nature. It's a key concept illustrating how certain transformations impact the common ratio but still maintain the sequence's geometric characteristic.

Common Difference in Sequences

The 'common difference' is a term that directly relates to both AP and GP sequences. In an AP, the common difference is the constant amount that each term increases by, making it a linear relationship. In a GP, although the term 'difference' isn't typically used, the equivalent concept is the 'common ratio', which represents the factor by which we multiply each term to arrive at the next.

Finding the common difference is essential in solving many problems involving APs, as demonstrated in the provided exercise where finding the common difference was a pivotal part of proving that a sequence was harmonic. The meticulous step-by-step approach to isolating this difference showcases its crucial role in understanding sequences.

Reciprocal Relations

Reciprocal relations involve taking the reciprocal of each term in a sequence. In our context with sequences, particularly when dealing with HP, the reciprocal relationship is vital. An HP is defined by its property that the reciprocals of its terms form an AP. This is because the harmonic mean between two numbers can be re-framed in terms of the arithmetic mean of their reciprocals.

The exercise demonstrates this concept by transforming a given HP into an AP to simplify the process of establishing the relationship between the sequences. By understanding reciprocal relations, we can maneuver between HP and AP and utilize known strategies for AP to tackle complex HP problems, as reflected in the exercise's solutions.