Harmonic Progression Concepts

1. HP Term Evaluation

• Formula: If (a_1,a_2,a_3,\dots) are in HP, then their reciprocals

$$\frac1{a_1},\frac1{a_2},\frac1{a_3},\dots$$

are in AP. If the reciprocal AP has first term (a) and common difference (d), then

$$\boxed{T_n=\frac{1}{a+(n-1)d}}.$$

Use when: Finding a particular term of an HP.

• Example: Find the 6th term of the HP

$$\frac13,\frac15,\frac17,\dots$$

Solution: Take reciprocals:

$$3,5,7,\dots$$

This is an AP with

$$a=3,\qquad d=2.$$

6th reciprocal:

$$T_6=3+(6-1)(2)=13$$

Therefore, 6th term of HP:

$$\boxed{\frac1{13}}$$

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2. Harmonic Mean of Two Numbers

• Formula: The harmonic mean of (a) and (b) is

$$\boxed{HM=\frac{2ab}{a+b}}.$$

Also,

$$\frac{2}{HM}=\frac1a+\frac1b.$$

Use when: A problem asks for the harmonic mean of two quantities, especially rates/speeds.

• Example: Find the harmonic mean of (4) and (12).

Solution:

$$HM=\frac{2(4)(12)}{4+12}$$ $$=\frac{96}{16}=6$$ $$\boxed{HM=6}$$

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3. Harmonic Mean of Multiple Numbers

• Formula: For (n) positive numbers (x_1,x_2,\dots,x_n),

$$\boxed{HM=\frac{n}{\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n}}}.$$

Use when: Finding the combined harmonic mean of 3 or more values.

• Example: Find the harmonic mean of (2,3,6).

Solution:

$$HM=\frac{3}{\frac12+\frac13+\frac16}$$ $$=\frac{3}{\frac{3+2+1}{6}} =\frac{3}{1}$$ $$\boxed{HM=3}$$

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4. Inserting Harmonic Means

• Formula: To insert (n) harmonic means between (a) and (b), take reciprocals:

$$\frac1a,\frac1{H_1},\dots,\frac1{H_n},\frac1b$$

and form an AP. Its common difference is

$$\boxed{d=\frac{\frac1b-\frac1a}{n+1}}.$$

Use when: Finding one or more harmonic means between two numbers.

• Example: Insert two harmonic means between (2) and (6).

Solution: Reciprocals form an AP:

$$\frac12,\frac1{H_1},\frac1{H_2},\frac16$$

There are 3 equal intervals:

$$d=\frac{\frac16-\frac12}{3} =\frac{-\frac13}{3} =-\frac19$$

Therefore,

$$\frac1{H_1}=\frac12-\frac19=\frac7{18}$$ $$H_1=\frac{18}{7}$$

And:

$$\frac1{H_2}=\frac7{18}-\frac19=\frac5{18}$$ $$H_2=\frac{18}{5}$$ $$\boxed{H_1=\frac{18}{7},\quad H_2=\frac{18}{5}}$$

HP:        2  ───── H₁ ───── H₂ ───── 6
Reciprocals:
          1/2 ──── 7/18 ─── 5/18 ─── 1/6
             ← common difference = -1/9 →

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5. HP Sum Using Reciprocal AP

• Formula: If the reciprocals of HP terms form an AP,

$$a_n=\frac1{A+(n-1)d},$$

then

$$\boxed{S_n=\sum_{k=0}^{n-1}\frac1{A+kd}}.$$

Use when: A question asks for the sum of a finite number of HP terms. Unlike AP, there is generally no simple standard sum formula.

• Example: Find the sum of the first 3 terms of

$$\frac12,\frac14,\frac16,\dots$$

Solution:

$$S_3=\frac12+\frac14+\frac16$$

Taking LCM (12):

$$S_3=\frac6{12}+\frac3{12}+\frac2{12} =\frac{11}{12}$$ $$\boxed{\frac{11}{12}}$$

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Advanced Variants

6. Relation Between AM, GM and HM

• Formula: For positive numbers (a,b),

$$\boxed{AM=\frac{a+b}{2}},\qquad \boxed{GM=\sqrt{ab}},\qquad \boxed{HM=\frac{2ab}{a+b}}$$

and

$$\boxed{AM\ge GM\ge HM}.$$

Also,

$$\boxed{GM^2=AM\times HM}.$$

Use when: A question gives one mean and asks for another.

• Example: The AM of two positive numbers is (10) and their HM is (6). Find their GM.

Solution: Using

$$GM^2=AM\times HM$$ $$GM^2=10\times6=60$$ $$\boxed{GM=2\sqrt{15}}$$

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7. HP and AP Parameter Problems

• Formula: If (x,y,z) are in HP, then

$$\boxed{\frac1x,\frac1y,\frac1z\text{ are in AP}}$$

so

$$\boxed{\frac2y=\frac1x+\frac1z}.$$

Equivalently,

$$\boxed{y=\frac{2xz}{x+z}}.$$

Use when: Three quantities are in HP and one term must be found.

• Example: If (4,x,12) are in HP, find (x).

Solution:

$$\frac2x=\frac14+\frac1{12}$$ $$\frac2x=\frac3{12}+\frac1{12}=\frac13$$ $$x=6$$ $$\boxed{x=6}$$

HP condition:

4 ───── x ───── 12
↑       ↑        ↑
1/4 ─── 1/x ─── 1/12
       AP