# Harmonic Sequence

## What is Harmonic Sequence?

We must first understand what Arithmetic Sequence or Arithmetic Progression is before learning about Harmonic Sequence or Harmonic Series. An arithmetic series is a set of numbers where the difference between any two consecutive elements is always constant. Arithmetic Progression, Geometric Progression, and Harmonic Progression are three forms of progression. Harmonic progression is a key concept in sequence and series, and it's closely related to arithmetic progression. In this article, we are going to discuss the harmonic sequence maths, harmonic progression in maths and its formula along with solved examples.

Harmonic sequence mathematics can be defined as The reciprocal form of the Arithmetic Sequence with numbers that can never be 0. The sum of harmonic sequence is known as Harmonic Series.

### Harmonic Progression in Maths

In Mathematics, we can define progression as a series of numbers arranged in a predictable pattern. It's a form of number set that adheres to strict, predetermined laws. The difference between the progression and a sequence is that a progression has a particular formula to compute its nth term, whereas a sequence is based on specific logical rules.

### Harmonic Progression Maths

A Harmonic Progression is a collection of real numbers that can be determined by multiplying the reciprocals of an arithmetic progression that doesn't contain zero. Any term in the series is treated as the harmonic mean of its two neighbours in harmonic progression. The series a, b, c, d,..., is an example of arithmetic progression. The harmonic progression can be written as $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$, $\frac{1}{d}$,... ...

### What is Harmonic Mean

The inverse of the arithmetic mean of the reciprocals is used to measure the harmonic mean. The following is the formula for calculating the harmonic mean:

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.

### First-term of Harmonic Progression

The first term of the Harmonic progression is denoted by a. The sum of the series can never be an integer except for the first term as it can be 1.

### Common Difference of Harmonic Progression

The common difference means the difference between any two-consecutive number in the series would be the same. This common difference is denoted as d.

### Example of First Term and Common Difference of Hp

If $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$ are three terms in Harmonic Progression then we can say that the first term is 1a and common difference

d = $\frac{1}{a}$ - $\frac{1}{b}$ = $\frac{1}{c}$ - $\frac{1}{b}$

Or $\frac{a-b}{ab}$ = $\frac{b-c}{bc}$

Or $\frac{a}{c}$ = $\frac{b-c}{a-b}$

### Harmonic Progression Formula

To find the sum of harmonic progression terms, we should find the corresponding arithmetic progression sum. It means that the nth term of the harmonic progression is equal to the reciprocal of the nth term of Arithmetic Progression. Thus, the formula to find the nth term of the harmonic progression series is given below:

n$^{th}$ term of the Harmonic Progression = $\frac{1}{[a+(n-1)]}$

Where

“a” is the first term of A.P

“d” is the common difference

“n” is the number of terms in A.P

The above formula can also be written as:

So, nth term of H.P = $\frac{1}{(n^{th} \text{ term of the corresponding A.P)}}$

### Sum of Harmonic Progression

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

Sum of n terms, S$_{n}$ = $\frac{1}{d}$ ln {$\frac{2a+(2n-1)d}{2a-d}$}

Where,

“a” is the first term of A.P

“d” is the common difference of A.P

“ln” is the natural logarithm

### Properties of Harmonic Progression

• No term of H.P. can be zero.

• If H is the H.M. between a and b, then we have the following properties

$\frac{1}{H-a}$ + $\frac{1}{H-b}$ = $\frac{1}{a}$ + $\frac{1}{b}$

(H - 2a)(H - 2b) = H$^{2}$

$\frac{H+a}{H-a}$ + $\frac{H+b}{H-b}$ = 2

• If a and b be two positive real numbers, then the relation between A.M, G.M and H.M is

A.M x H.M = G.M$^{2}$

### Solved Examples:

1. Find the 4$^{th}$ and 8$^{th}$ Terms of the Given Harmonic Progression 6, 4, 3,…

Sol: Given H.P terms are 6, 4, 3

Now, convert it into the arithmetic progression from the given H.P

A.P term will be ⅙, ¼, ⅓, ….

Here, T$_{2}$ - T$_{1}$ = T$_{3}$ - T$_{2}$ = $\frac{1}{4}$ - $\frac{1}{6}$ = $\frac{1}{12}$ = d

Now, we will find the 4th term of an A.P, using the formula of nth term,

The nth term of an A.P = a+(n-1)d

Here, a = $\frac{1}{6}$ d = $\frac{1}{12}$

Now, we have to find the 4$^{th}$ term.

So, take n = 4

Now put the values in the formula.

4th term of an A.P = $\frac{1}{6}$ + (4 - 1) $\frac{1}{12}$

= $\frac{1}{6}$ + $\frac{3}{12}$

= $\frac{1}{6}$ + $\frac{1}{4}$

= $\frac{5}{12}$

Similarly,

8th term of an A.P = = $\frac{1}{6}$ + (8 - 1) $\frac{1}{12}$

= $\frac{1}{6}$ + $\frac{7}{12}$

= $\frac{9}{12}$

Since H.P is the reciprocal of terms in A.P, therefore we can write the values as:

4th term of an H.P is $\frac{1}{4^{th} \text{ term of on A.P}}$ = $\frac{12}{5}$

8th term of an H.P is $\frac{1}{8^{th} \text{ term of on A.P}}$ = $\frac{12}{9}$ = $\frac{4}{3}$

2. Find the Sum of the Below Harmonic Sequence.

1/12 + 1/24 + 1/36 +1/48 +1/60

Sol: As we know the reciprocal of H.P terms are in A.P

So A.P term will be 12,24,36,48,60

Here a= 12, d=24-12=12, n=5

Now put the above values in the sum of A.P formula is

S$_{n}$ = $\frac{n}{2}${2a + (n - 1)d}

So put the value of a,d and n in the above equation

S$_{5}$ = $\frac{5}{2} (2 \times 12 + (5 - 1)12)$

= $\frac{5}{2} (24 + 48)$

= 180

Sum of H.P terms will be reciprocal of A.P

So, the sum of 5 terms of H.P will be $\frac{1}{180}$