In mathematics, a harmonic progression (HP or harmonic sequence HS) is a progression formed by taking the reciprocals of an arithmetic progression.

The sequence 1,2,3,4,5,6,... is an arithmetic progression, so its reciprocals

\( \frac{1}{1}, \frac{1}{2},
\frac{1}{3}, \frac{1}{4}, ... \) are a harmonic progression.

If each term of an harmonic sequence is multiplied or divide by a constant, the sequence of the resulting number are also harmonic sequence. For example,

- if a,b,c,d,... are harmonic sequence
- then ka,kb,kc,kd,... are harmonic sequence where k is constant and \(k \ne 0\)
- \(\frac{a}{k},\frac{b}{k},\frac{c}{k},\frac{d}{k},... \) are harmonic sequence where \(k \ne 0\)

#### General term of Harmonic sequence

Let a,a+d,a+2d,a+3d, .... are the terms of arithemetic sequence

Then

terms of Harmonic sequence are given by

\( \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d}+\frac{1}{a+3d},...\)

Now,

The nth term of the sequence is given by

\( a_n=\frac{1}{a+(n-1)d}\)

#### 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 =\( \frac{n}{(\frac{1}{a})+(\frac{1}{b})+(\frac{1}{c})+(\frac{1}{d})+... } \)

Where a,b,c,d are the values, and n is the number of values present.

#### Sum of Harmonic Series

If \( \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d}\) is given harmonic progression, the formula to find the sum of n terms in the harmonic progression is given by the formula:

Sum of nth terms, \( S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]\)

where,

“a” is the first term of A.P

“d” is the common difference of A.P

“ln” is the natural logarithm

Proof

Given \( \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d}\) are in harmonic progression,

We set the values as follows,

a=1,d=1

Then, \( \frac{1}{1},\frac{1}{1+1},\frac{1}{1+2.1},…,\frac{1}{1+(n-1)1}\) are the terms

\( 1,\frac{1}{2},\frac{1}{3}, ...,\)are the terms

Now, the Riemann sum of the function \( f(x)=\frac{1}{x}\) approximates the sum of the harmonic series given above

Therefore,

We set the values as follows,

\( S_n= \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx \)

For any common difference,the formula becoms

\( S_n=\frac{1}{d} \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx \)

or \( S_n= \frac{1}{d} \int_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx \)

or \( S_n= \frac{1}{d} \left [ \ln x \right]_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d} \)

or \( S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]\)

Therefore

Sum of nth terms, \( S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n–1)d}{2a–d} \right ]\)

#### A.M, G.M, H.M and their relations

A finite sequence consisting ore than two terms has one or more terms in between the first and last terms. Theses between terms are called means of the sequence. Precisely

- if a,b,c,d are in arithmetic sequence then b,c are arithmetic means (AM)
- item if a,b,c,d are in geometric sequence then b,c are geometric means (GM)
- if a,b,c,d are in harmonic sequence then b,c are harmonic means (HM)

#### Theorem 1

Given any two numbers a and b the AM, GM, and HM are as follows.

- AM=\(\frac{a+b}{2}\)
- GM=\(\sqrt{ab}\)
- HM=\(\frac{2ab}{a+b}\)

The proof are as follows:

- Let AM is the single mean between a and b, then

AM-a=b-AM

or 2AM=a+b

or AM=\(\frac{a+b}{2}\)

- Let GM is the single mean between a and b, then

\( \frac{GM}{a}=\frac{b}{GM}\)

or \(GM^2=ab \)

or \( GM=\sqrt{ab}\)

- Let HM is the single mean between a and b, then

\( \frac{1}{HM}-\frac{1}{a}=\frac{1}{b}-\frac{1}{HM}\)

or \( \frac{2}{HM}=\frac{1}{a}+\frac{1}{b}\)

or \( HM=\frac{2ab}{a+b}\)

#### Theorem 2

Let a and b are two non-negative numbers, then

- \( GM^2=AM \times HM \)
- \( AM \ge GM \ge HM \) Arithmetic mean is greater than geometric mean and harmonic mean, and geometric mean is greater than harmonic mean.

Let a and b are two non-negative numbers then,

\( AM=\frac{a+b}{2}, GM=\sqrt{ab}, HM=\frac{2ab}{a+b}\)

Now, we have

\( GM^2=ab\)

or
\( GM^2=\frac{a+b}{2} \times \frac{2ab}{a+b} \)

or
\( GM^2=AM \times HM \)

For the second part, we write

\(AM-GM=\frac{a+b}{2}-\sqrt{ab}\)

or
\(AM-GM=\frac{a+b-2\sqrt{ab}}{2}\)

or
\(AM-GM=\frac{{{\sqrt{a}}^{2}}+{{\sqrt{b}}^{2}}-2\sqrt{a}\sqrt{b}}{2}\)

or
\(AM-GM=\frac{{{( \sqrt{a}-\sqrt{b} )}^{2}}}{2}\)

or
\(AM\ge GM\) (1)

Similarly,

or
\(GM-HM=\sqrt{ab}-\frac{2ab}{a+b}\)

or
\(GM-HM=\frac{\sqrt{ab}( a+b )-2ab}{a+b}\)

or
\(GM-HM=\frac{\sqrt{ab}( a+b )-2\sqrt{ab}\sqrt{ab}}{a+b}\)

or
\(GM-HM=\frac{\sqrt{ab}}{a+b}( a+b-2\sqrt{ab} ) \)

or
\(GM-HM=\frac{\sqrt{ab}}{a+b}{{( \sqrt{a}-\sqrt{b} )}^{2}}\)

or
\(GM\ge HM\) (2)

Combining (1) and (2), we get

or
\(AM\ge GM\ge HM\)

#### Visualization of the proof

Let us suppose that a and b are two given numbers. Now, draw a semi circle with diameter a+b.

Then OP is the arithemetic mean given by

\( AM =\frac{a+b}{2} \)

#### Visualization of AM

#### Visualization of GM

#### Visualization of HM

#### Visualization of AM,GM and HM

By the property of radius and diameter, we get that

\( AM =\frac{a+b}{2} \)

Also, by the mean proportionality property (squaring a rectangle), we can obtain by using the property of similarity that, DQ is the geometric mean given by

\( GM =\sqrt{ab} \)

By using proportionality, we get

Triangle ADQ and QDB are similar with AD=a, DB=b, so we have

\( \frac{GM}{a}=\frac{b}{GM} \)

or \( GM= \sqrt{ab} \)

Again, by using the property of similarity on OCDE, we get that, QR is the harmonic mean given by

\( HM =\frac{2ab}{a+b} \)

By using proportionality, we get

Triangle DRQ and ODQ are similar with QR=GM,QD=\(\sqrt{ab}\), OD=\(\frac{a-b}{2}\), so we have

\( \frac{HM}{\sqrt{ab}}=\frac{\sqrt{ab}}{\frac{a+b}{2}} \)

or \( HM= \frac{2ab}{a+b} \)

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