# Sum of Harmonic Progression Series

\[\frac{1}{d} In \frac{2a+(2n-1)d}{2a-d}\]

## Sum of Harmonic Progression Series

The sum of a Harmonic Progression (HP) series is the sum of the reciprocals of the terms of an Arithmetic Progression (AP). There is no simple formula for the sum of an HP series like there is for an AP or GP, but it is often expressed as the sum of individual terms: \(S = \sum_{k=1}^{n} \frac{1}{a + (k-1)d}\).

#### Introduction to Harmonic Progression (HP)

A Harmonic Progression (HP) is a sequence of numbers where the reciprocals of the terms form an Arithmetic Progression (AP). In other words, if you take the reciprocal of each term in an HP, you get a series in AP.

#### Definition

If \( a, a+d, a+2d, \ldots \) is an AP, then the corresponding HP is:

\[

\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \ldots

\]

#### Sum of Harmonic Progression Series

Unlike Arithmetic and Geometric Progressions, there is no straightforward formula for the sum of an HP series. Instead, the sum is usually expressed in terms of the reciprocals of the AP terms.

#### Calculating the Sum

To find the sum of the first \( n \) terms of an HP series, you add the reciprocals of the corresponding AP terms. If the AP series is given by \( a, a+d, a+2d, \ldots \), the HP series sum \( S_n \) is:

\[

S_n = \sum_{k=1}^{n} \frac{1}{a + (k-1)d}

\]

#### Example

Let’s find the sum of the first 4 terms of the HP corresponding to the AP: \( 2, 4, 6, 8 \).

1. AP Series: \( 2, 4, 6, 8 \)

2. HP Series: \( \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8} \)

#### Step-by-Step Calculation

\[

S_4 = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8}

\]

Finding a common denominator (24) to add these fractions:

\[

S_4 = \frac{12}{24} + \frac{6}{24} + \frac{4}{24} + \frac{3}{24}

\]

\[

S_4 = \frac{12 + 6 + 4 + 3}{24} = \frac{25}{24}

\]

So, the sum of the first 4 terms of the HP series is:

\[

S_4 = \frac{25}{24}

\]