Problem 28
Question
Insert two harmonic means between \(\frac{7}{9}\) and \(\frac{7}{15}\)
Step-by-Step Solution
Verified Answer
The two harmonic means between \(\frac{7}{9}\) and \(\frac{7}{15}\) are \(\frac{7}{11}\) and \(\frac{7}{13}\).
1Step 1: Understanding Harmonic Means
The harmonic mean between two numbers is a type of average, often used for rates or ratios. When inserting two harmonic means, A and B, between two numbers, X and Y, it means that X, A, B, and Y will form a harmonic progression (HP). The reciprocal of the harmonic mean of reciprocals of X and Y will give the value of each harmonic mean.
2Step 2: Reciprocal of the Given Numbers
Find the reciprocal of each of the given numbers. The reciprocal of a number is 1 divided by that number. So, the reciprocal of \(\frac{7}{9}\) is \(\frac{9}{7}\) and the reciprocal of \(\frac{7}{15}\) is \(\frac{15}{7}\).
3Step 3: Determine the Common Difference
In a harmonic progression, the difference between the reciprocals of successive terms is constant. This difference is called the common difference (d). To find d, subtract the reciprocals of the two given numbers and divide by the number of desired means plus one. \(d = \frac{\frac{15}{7} - \frac{9}{7}}{2+1} = \frac{6}{7} \div 3 = \frac{2}{7}\)
4Step 4: Finding Reciprocals of Harmonic Means
To find the reciprocals of the harmonic means A and B, add the common difference to the reciprocal of the first number for A, and add the common difference twice to the reciprocal of the first number for B. The reciprocals of harmonic means A and B will be:\[ \text{Reciprocal of A} = \frac{9}{7} + \frac{2}{7} = \frac{11}{7} \] \[ \text{Reciprocal of B} = \frac{11}{7} + \frac{2}{7} = \frac{13}{7} \]
5Step 5: Calculating Harmonic Means
Now, take the reciprocal of \(\frac{11}{7}\) and \(\frac{13}{7}\) to find the harmonic means A and B: \[ A = \frac{7}{11} \] \[ B = \frac{7}{13} \]
Key Concepts
Harmonic ProgressionReciprocal of a NumberCommon Difference
Harmonic Progression
When we talk about sequences in mathematics, one particular type that comes up is a harmonic progression (HP). This is a sequence of numbers where the reciprocals of the terms form an arithmetic progression (AP). To clarify, if you take the reciprocals of each term in a harmonic progression, you would get an evenly spaced sequence - this spaced sequence is known as an AP.
For example, if you have a harmonic progression such as \( 1, \frac{1}{2}, \frac{1}{3}, ... \), their reciprocals \( 1, 2, 3, ... \) represent an arithmetic progression with a common difference of 1. Understanding this concept is essential when inserting harmonic means between two given numbers, as this essentially means adding terms to the HP such that the entire sequence remains in harmony - the reciprocity with an arithmetic sequence is maintained.
In applications, HP is particularly useful in fields like physics and engineering, where it can be applied to understand phenomena that are inversely related, such as resistance in parallel circuits or average speed for a round trip.
For example, if you have a harmonic progression such as \( 1, \frac{1}{2}, \frac{1}{3}, ... \), their reciprocals \( 1, 2, 3, ... \) represent an arithmetic progression with a common difference of 1. Understanding this concept is essential when inserting harmonic means between two given numbers, as this essentially means adding terms to the HP such that the entire sequence remains in harmony - the reciprocity with an arithmetic sequence is maintained.
In applications, HP is particularly useful in fields like physics and engineering, where it can be applied to understand phenomena that are inversely related, such as resistance in parallel circuits or average speed for a round trip.
Reciprocal of a Number
The reciprocal of a number is simply one divided by that number. It's an essential concept when dealing with fractions, harmonics, and various forms of equations. If you have whole number, its reciprocal is \( \frac{1}{\text{number}} \). For a fraction, like \( \frac{a}{b} \), the reciprocal is \( \frac{b}{a} \). It might seem simple, but reciprocals are fundamental in the context of harmonic progressions.
Understanding reciprocals provides insight into how numbers or terms relate to each other within a sequence. They serve as the 'flip-side' and will often reveal proportional relationships that are not apparent with the original numbers. Reciprocals are crucial when computing values such as harmonic means because you first need to work with the reciprocals of the numbers involved, then flip the result again to obtain the actual mean in the original context.
Understanding reciprocals provides insight into how numbers or terms relate to each other within a sequence. They serve as the 'flip-side' and will often reveal proportional relationships that are not apparent with the original numbers. Reciprocals are crucial when computing values such as harmonic means because you first need to work with the reciprocals of the numbers involved, then flip the result again to obtain the actual mean in the original context.
Common Difference
The term 'common difference' usually arises when discussing arithmetic progressions (AP), where it refers to the constant difference between successive terms. However, when we translate this to harmonic progressions, the concept adapts to refer to the difference between the reciprocals of the successive terms of the harmonic progression. The common difference is symbolized as 'd' and is a critical aspect of identifying or establishing a sequence.
To find the common difference in a harmonic progression, as we've seen in the textbook solution, you need to subtract the two given reciprocals and then divide by the number of terms you wish to insert plus one. This process stems from the fact that adding harmonic means expands the reciprocal AP, and thus, to maintain uniform spacing, 'd' has to be calculated this way. The concept of common difference enables us to construct desired sequences and calculate measures such as harmonic means, making it a powerful tool for solving related mathematical problems.
To find the common difference in a harmonic progression, as we've seen in the textbook solution, you need to subtract the two given reciprocals and then divide by the number of terms you wish to insert plus one. This process stems from the fact that adding harmonic means expands the reciprocal AP, and thus, to maintain uniform spacing, 'd' has to be calculated this way. The concept of common difference enables us to construct desired sequences and calculate measures such as harmonic means, making it a powerful tool for solving related mathematical problems.
Other exercises in this chapter
Problem 27
Show that the harmonic mean between two numbers \(a\) and \(b\) is given by $$\text { Harmonic Mean }=\frac{2 a b}{a+b}$$
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Verify the first four terms of each infinite binomial series. $$(1+5 a)^{-5}=1-25 a+375 a^{2}-4375 a^{3} \dots$$
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