Question

If the second and the seventh terms of a Harmonic Progression are \(\frac{1}{5}\) and \(\frac{1}{25}\), then find the series. (1) \(1, \frac{1}{5}, \frac{1}{9}, \ldots \ldots \ldots\) (2) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \ldots \ldots\) (3) \(\frac{1}{7}, \frac{1}{5}, \frac{1}{3}, \ldots \ldots .\) (4) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots \ldots . .\)

Short answer

Given the second and seventh terms of a Harmonic Progression (HP), we found that the series the HP belongs to is (1) 1, 1/5, 1/9, .... We did this by converting the HP to an Arithmetic Progression (AP), finding the AP's first term and common difference using the given conditions, and then using these values to identify which of the choices matched the given HP.

Step-by-step solution

Convert the HP to an AP

We are given the second (\(\frac{1}{5}\)) and the seventh (\(\frac{1}{25}\)) terms of an HP. Let's consider their corresponding terms in an AP. Say, \(\frac{1}{a_1 + d} = \frac{1}{5}\) (second term) And, \(\frac{1}{a_1 + 6d} = \frac{1}{25}\) (seventh term) We have then the arithmetic sequence \(a_1 + d, a_1 + 6d\).

Find the first term a_1 and the common difference d

Using the given terms in AP, we can write the following system of equations: \((1)\) \(a_1 + d = 5\) \((2)\) \(a_1 + 6d = 25\) Subtract equation \((1)\) from equation \((2)\) to eliminate \(a_1\): \(d = \frac{(25-5)}{(6-1)} = \frac{20}{5} = 4\) Now, use the value of d to find the first term a_1: \(a_1 = 5 - d = 5 - 4 = 1\) So, the first term \(a_1 = 1\) and the common difference \(d = 4\).

Identify the HP series

Now that we have the values of \(a_1 = 1\) and \(d = 4\), we can find the HP series by comparing the given choices: - \(\textrm{(1)} 1, \frac{1}{5}, \frac{1}{9}, \ldots \ldots \ldots\) - \(\textrm{(2)} \frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \ldots \ldots\) - \(\textrm{(3)} \frac{1}{7}, \frac{1}{5}, \frac{1}{3}, \ldots \ldots .\) - \(\textrm{(4)} \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots \ldots . .\) Now let's go through the choices one by one and look for an HP with \(a_1 = 1\) and \(d = 4\): 1. In the first choice, the HP is \(1, \frac{1}{5}, \frac{1}{9},\ldots\) and the corresponding AP would be \(1,5,9,\ldots\) which satisfies our conditions (\(a_1 = 1\) and \(d = 4\)), so this is the correct HP. The correct answer to the problem is \(\boxed{\textrm{(1)}\ 1, \frac{1}{5}, \frac{1}{9}, \ldots \ldots \ldots}\).

Key concepts

Arithmetic Progression

Understanding an Arithmetic Progression (AP) is the foundational step when working with series and sequences, particularly when one encounters a Harmonic Progression (HP). An AP is a sequence of numbers in which each term after the first is obtained by adding a fixed, non-zero number, known as the common difference, to the previous term. This consistent increment forms a linear sequence.

The simplest example of an AP is the natural numbers themselves: 1, 2, 3, 4, 5, and so on, where the common difference is 1. One can write down the general formula for the nth term of the AP as: \[ a_n = a_1 + (n - 1)d \] where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the position of the term within the sequence.
When it comes to solving problems involving AP, it's common to leverage the characteristics of the sequence, like using the formula for the nth term or the sum of the first n terms.

Sequence and Series

In mathematics, the concept of sequence and series pertains to the orderly collection of numbers defined by a particular rule. A sequence is simply a list of objects (normally numbers) that follow a certain pattern. The individual numbers in the list are called terms.

A series is the sum of the elements of a sequence. It’s important to distinguish between the sequence itself - which is the list, and the series - which is the sum. For example, the sequence 2, 4, 6, 8, ..., has its corresponding arithmetic series as 2 + 4 + 6 + 8 + .... Sequences can be finite or infinite, and the same applies to series.

• Arithmetic Sequence: Numbers separated by a common difference.
• Geometric Sequence: Numbers separated by a common ratio.
• Harmonic Sequence: Reciprocals of an arithmetic sequence.

When solving for sequences of numbers or converting between types, such as turning a HP into an AP, it is critical to understand the underlying formula that governs the pattern of numbers.

Common Difference

The common difference is a critical element of an Arithmetic Progression, functioning as the 'stepping stone' between each term. It is the constant value added to each term to get the next one in the AP. In the sequence 5, 8, 11, 14, ..., the common difference is 3, because we add 3 to one term to get to the next.

To find the common difference when given certain terms of an AP, you subtract one term from another. This is particularly useful when looking to identify an unknown AP or HP, which is derived from an AP. In our exercise, we found the common difference \(d\) by subtracting the second given AP term from the seventh term and dividing by 5 (the number of terms between them).

By mastering the concept of common difference, students can not only find terms in an AP but also solve complex problems involving other types of sequences, such as HPs, by first converting them into APs and then applying their properties. Recognizing and understanding the common difference aids in piecing together the arithmetic puzzle inherent in numerical sequences.