Progressions are numbers arranged in a particular sequence such that they form a predictable order. A progression has a specific formula to compute its nth term, whereas a sequence is based on specific logical rules. A progression is distinguished into three types: Arithmetic Progression, Geometric Progression and Harmonic Progression.

Any term in the sequence is considered as the harmonic mean of its two consecutive terms, in harmonic progression. In this article, we will provide detailed information about harmonic progression. Continue reading to find out more!

Define Harmonic Progression

A harmonic progression (H.P.) is defined as a sequence of real numbers determined by taking the reciprocals of the arithmetic progression that do not contain \(0\). In harmonic progression, any term in the series is considered the harmonic means of its two neighbors.
For example, if the sequence \(a, b, c, d,….\) is viewed as an arithmetic progression; then the harmonic progression can be written as:
\(\frac{1}{a},\,\frac{1}{b},\,\frac{1}{c},\,\frac{1}{d},….\)

Harmonic Mean: Harmonic mean is finding as to the reciprocal of the arithmetic mean of the reciprocals. The formula to calculate the harmonic mean is given by:
Harmonic Mean \( = \frac{n}{{\left[ {\left( {\frac{1}{a}} \right) + \left( {\frac{1}{b}} \right) + \left( {\frac{1}{c}} \right) + \left( {\frac{1}{d}} \right) + \cdots } \right]}}\)
Where \(a, b, c, d\) are the values, and \(n\) is the number of values present.

Examples of Harmonic Progression

Some of the examples of harmonic progression are:

1. \(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},……\)
2. \(\frac{1}{5},\frac{1}{10},\frac{1}{15},\frac{1}{20},……\)
3. \(\frac{1}{4},\frac{1}{7},\frac{1}{10},\frac{1}{13},……\)
4. \(1, \frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},……\)

Harmonic Progression Method

To solve the harmonic progression problems, first, we should find the corresponding arithmetic progression sum. It means that the \({n^{{\rm{th}}}}\) term of the harmonic progression is equal to the reciprocal of the nth term of the corresponding A.P.
Example: \(1, \frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},……\)
Sequence \(1, \frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9},……\) is a harmonic progression (H.P). Because the sequence \(1, 3, 5, 7, 9,….\) is an arithmetic progression (A.P).

Harmonic Progression Formula

To solve the harmonic progression problems, first, we should find the corresponding arithmetic progression sum. It means that the \(n^{th}\) term of the harmonic progression is equal to the reciprocal of the \(n^{th}\) term of the corresponding A.P. Thus, the formula to find the \(n^{th}\) term of the harmonic progression series is given as:
The \(n^{th}\) term of the harmonic progression (H.P) \( = \frac{1}{{\left[{a + \left({n – 1} \right)d} \right]}}\)
where,
“\(a\)” is the first term of A.P
“\(a\)” is a common difference.
“\(n\)” is the number of terms in A.P
The above formula can also be written as:
The nth term of H.P \( = \frac{1}{{\left({{{n^{th} }}\,{\text{term}}\,{\text{of}}\,{\text{the}}\,{\text{corresponding}}\,{\text{A}}{\text{.P}}} \right)}}\)

Harmonic Progression Sum

If \(\frac{1}{a},\frac{1}{ {a + d}},\frac{1}{ {a + 2d}}, \ldots ,\frac{1}{ {a + (n – 1)d}}\) is given harmonic progression, the formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of \(n^{th}\) terms, \({S_n} = \frac{1}{d}\ln \left[{\frac{{2a + \left({2n – 1} \right)d}}{{2a – d}}}\right]\)
where,
“\(a\)” is the first term of A.P
“\(d\)” is the common difference of A.P
“\(ln\)” is the natural logarithm

Relation Between Arithmetic Mean, Geometric Mean, and Harmonic Mean

For any two numbers, if \(A.M, G.M, H.M\) are the arithmetic, geometric, and harmonic Mean respectively, then the relationship between these three is given by:
1. \(G.M^2 = A.M \times H.M\), where \(A.M, G.M, H.M\) are in \(G.P\)
2. \(A.M ≥ G.M ≥ H.M\)

Applications of Harmonic Progression

Some of the applications of harmonic progression are:

1. In everyday life, the harmonic formulae can also be used by scientists to conclude the value of their experiments.
2. Harmonic progression is used to establish how water boils each time the temperature is changed with the same value.
3. The notes used in music uses the concept of harmonic sequence.
4. Harmonic progression is used to measure the number of raindrops – calculating the amount of rainfall creates the illusion that it could be estimated when the series is infinite.
5. Traffic bunches – it is also a real-life example of a harmonic sequence.

Solved Examples on Harmonic Progression

Q.1. Determine the \(5^{th}\) and \(8^{th}\) term of the harmonic progression \(6, 4, 3,….\)
Ans: Given: H.P \(= 6, 4, 3\)
Now, let us consider and take the arithmetic progression from the given H.P
A.P \( = \frac{1}{6},\,\frac{1}{4},\,\frac{1}{3},……\)
Here, \(T_2 – T_1 = T_3 – T_2 = \frac{1}{12} = d\)
So, to find the \(5^{th}\) term of an A.P, use the formula,
The \(n^{th}\) term of an A.P \(= a + (n – 1) d\)
Here, \(a = \frac{1}{6}\), \(d = \frac{1}{12}\)
Now, we must find the \(5^{th}\) term.
So, take \(n = 5\)
Now put the values in the formula.
\(5^{th}\) term of an A.P \(= \frac{1}{6} + (5 – 1) \frac{1}{12}\)
\( \Rightarrow \frac{1}{6} + 4 \times \frac{1}{{12}}\)
\( \Rightarrow \frac{1}{6} + \frac{1}{3}\)
\( \Rightarrow \frac{1 + 2}{6} = \frac{1}{2}\)
So, \(5^{th}\) term of an A.P is \(\frac{1}{2}\).
Similarly,
\(8^{th}\) term of an A.P \(= \frac{1}{6} + (8 – 1) \frac{1}{12}\)
\( \Rightarrow \frac{1}{6} + 7 \times \frac{1}{{12}}\)
\( \Rightarrow \frac{1}{6} + \frac{7}{12}\)
\( \Rightarrow \frac{2 + 7}{12} = \frac{9}{12} = \frac{3}{4}\)
We know Harmonic Progression is the reciprocal of an Arithmetic progression (A.P), we can write the values as:
\(5^{th}\) term of an H.P \(=\) reciprocal of \(5^{th}\) term of an A.P is \(\frac{2}{1} = 2\)
\(8^{th}\) term of an H.P \(=\) reciprocal of \(8^{th}\) term of an A.P is \(\frac{4}{3}.\)

Q.2. Is this sequence \(1,\,\frac{1}{4},\,\frac{1}{7},\,\frac{1}{{10}},\,\frac{1}{{13}},……\) a harmonic progression?
Ans: Sequence \(1,\,\frac{1}{4},\,\frac{1}{7},\,\frac{1}{{10}},\,\frac{1}{{13}},……\) is a harmonic progression (H.P). Because the sequence \(1, 4, 7, 10, 13….\) is an arithmetic progression (A.P).

Q.3. Find the \(16^{th}\) term of H.P. if the \(6^{th}\) and \(11^{th}\) term of H.P. are \(10\) and \(18\), respectively.
Ans: The H.P is written in terms of A.P are given below:
\(6^{th}\) term of A.P \( = a + 5d = \frac{1}{{10}}\)……..(1)
\(11^{th}\) term of A.P \( = a + 10d = \frac{1}{{18}}\)……..(2)
By solving these two equations, we get
\(a = \frac{13}{90}\), and \(d = \frac{-2}{225}\)
To find the \(16^{th}\) term of the A.P., we can write the expression in the form,
\(a + 5d = \left({\frac{{13}}{{90}}}\right) – \left({\frac{2}{{15}}} \right) = \frac{1}{{90}}\)
Thus, the \(16^{th}\) term of an H.P \(=\) reciprocal of \(16^{th}\) term of an A.P \(= 90\)
Therefore, the obtained answer is \(90\).

Q.4. If the sum of first \(11\) terms of an A.P. series is \(110\), find the \(6^{th}\) term of the corresponding H.P.
Ans: Reciprocals of the first \(11\) terms will be A.P.
Therefore, \({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
\(S_n = 110\)
\(n = 11\)
So,we have, \(110 = \frac{11}{2} \left[{2a + \left({11 – 1} \right)d} \right]\)
\( \Rightarrow 110 = \frac{{11}}{2}\left[{2a + 10d} \right]\)
\( \Rightarrow 220 = 22a + 110d\)
\( \Rightarrow 22a + 110d = 220\)
\( \Rightarrow a + 5d = 10\)
which is the \(6^{th}\) term of the A.P series
Therefore, the 6th term in harmonic progression is \(\frac{{1}}{10}\)

Q.5. The sum of first \(11\) terms of an A.P. series is \(220\), then find the \(6^{th}\) term of the corresponding H.P.
Ans: Reciprocals of the first \(11\) terms will be A.P.
Therefore, \({S_n} = \frac{n}{2}\left[{2a + \left({n – 1} \right)d} \right]\)
\(S_n = 220\)
\(n = 10\)
\( \Rightarrow 220 = \frac{{11}}{2}\left[{2a + (11-d)d} \right]\)
\( \Rightarrow 440 = {11}\left[{2a + 10d} \right]\)
\( \Rightarrow 440 = 22a + 110d\)
\( \Rightarrow 22a + 110d = 440\)
\( \Rightarrow a + 5d = 20\)
which is the \(6^{th}\) term of the A.P series
Therefore, the \(6^{th}\) term in harmonic progression is \(\frac{{1}}{20}\)

Summary

A harmonic progression is a sequence of finding real numbers by taking the reciprocals of the arithmetic progression that does not contain \(0\). This article helps to understand in detail harmonic progression, its formulas, solving methods. Harmonic Progression has various applications in our daily life. Harmonic progression is helpful in understanding how water boils each time the temperature changes with the same value. It is also used to measure the number of raindrops.

FAQs on Harmonic Progression

Q.1. What is the condition for harmonic progression?
Ans: The condition for harmonic progression is if a series of terms is known as a harmonic progression series when the reciprocals of elements are in arithmetic progression.

Q.2. What is a harmonic sequence and examples?
Ans: A harmonic sequence is the sequence of real numbers determined by taking the reciprocals of the arithmetic progression that does not contain \(0\). In harmonic progression, any term in the series is considered the harmonic means of its two neighbors.
For example, if the sequence \(p, q, r, s,…..\) is an arithmetic progression; the harmonic progression can be written as
\(\frac{1}{p},\,\frac{1}{q},\,\frac{1}{r},\,\frac{1}{s},…….\)
Examples:
1. If \(4, 8, 12, 16,…..\) are in A.P, then \(\frac{1}{4},\,\frac{1}{8},\,\frac{1}{12},\,\frac{1}{16},…….\) are in H.P.
2. If \(10, 12, 14, 16\) are in A.P, then \(\frac{1}{10},\,\frac{1}{12},\,\frac{1}{14},\,\frac{1}{16},…….\) are in H.P.

Q.3. How do you know if a sequence is harmonic?
Ans: A harmonic progression (H.P.) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression that does not contain \(0\). In a harmonic progression, any term in the series is considered as the harmonic means of its two neighbors.

Q.4. What is the harmonic progression formula?
Ans: It means that the nth term of the harmonic progression is equal to the reciprocal of the nth term of the corresponding A.P. Thus, the formula to find the nth term of the harmonic progression series is given as:
The \(n^{th}\) term of the harmonic progression (H.P) \( = \frac{1}{{\left[{a + \left({n – 1} \right)d} \right]}}.\)

Q.5. What is the formula to find the sum of harmonic progression?
Ans: If \(\frac{1}{a},\frac{1}{{a + d}},\frac{1}{{a + 2d}}, \ldots ,\frac{1}{{a + (n – 1)d}}\) is given harmonic progression, the formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of \(n^{th}\) terms, \({S_n} = \frac{1}{d}\ln \left[{\frac{{2a + \left({2n – 1} \right)d}}{{2a – d}}} \right]\)
Where,
“\(a\)” is the first term of A.P
“\(d\)” is the common difference of A.P
“\(ln\)” is the natural logarithm.

Q.6. Write the application of harmonic progression in everyday life.
Ans: Some of the applications of harmonic progression are:
1. In everyday life, the harmonic formulae can also be used by scientists to conclude the value of their experiments.
2. Harmonic progression is used to establish how water boils each time the temperature is changed with the same value.
3. The notes which are used in music uses the concept of harmonic sequence.
4. Harmonic progression is used to measure the number of raindrops by calculating the amount of rainfall creates the illusion that it could be estimated when the series is infinite.
5. Traffic bunches – It is also a real-life example of a harmonic sequence.

We hope this detailed article on harmonic progression helped you in your studies. If you have any doubts or queries regarding this topic, feel to ask us in the comment section, and we will assist you at the earliest.