The Harmonic Progression is a part of the progression topic from CAT quantitative aptitude section. Learn the detailed concepts of the topic with easily-understandable explanations and solved illustrations.

**Introduction: Harmonic Progression**

Numbers a, b, c are said to be in Harmonic Progression HP, if the reciprocals of these numbers (1/a,1/b,1/c) are in an AP. Also, no term in an HP, can be zero.

**Example:** Take the numbers: 1, 1/3 and 1/5.

Now, the reciprocals of the given numbers are 1, 3, 5 which are in AP; so, it can be said that 1, 1/3, 1/5 are in HP.

**General form Of an HP**

**The n ^{th} term of an HP = 1/((n^{th} term of corresponding AP))**

Example: take the HP: 1/2, 1/5, 1/8

The corresponding AP is 2, 5, 8 with first term 2 and common difference = 3.

The 6^{th} term in this AP = 2 + (5)(3) = 17.

The corresponding 6th term in the HP = 1/17.

**Example Question:**

**If the first two terms in an HP are 24 and 12, then find the 4 ^{th} term?**

**Solution:**

Let 3^{rd} term = x

Since the reciprocals are in AP, 1/24,1/12,1/x are in AP

i.e 1/x-1/12=1/12-1/24 1/x=1/6–1/24=1/8

Let the 4^{th} term = y 1/y-1/8=1/8-1/12=>1/y=1/6=>y =6

Required answer, 4^{th} term = 6

Alternately, if you take the percentage equivalents, it becomes easier as,

1/24= 4.16

1/12= 8.33. 4.16 & 8.33 are the first two terms in the AP,

3^{rd} term = 12.5, 4^{th} term = 16.66 = 1/6,

which implies that the 4^{th} term in the HP = 6.

**Harmonic Mean (HM):**

Let a and b be two given quantities. It is required to insert n harmonic means h1, h2, h3,….h_{n }between the quantities a and b

=> 1/a,1/h_{1},1/h_{2},…..1/h_{n},1/b are in A.P

Harmonic mean of n numbers a_{1},a_{2},……an is given by: n/[1/a_{1} + 1/a_{2} + 1/a_{3} +?.1/a_{n} ]

**Harmonic Mean Formula:**

If a, b, c are in Harmonic Progression, then

**HM, b =2/[1/a + 1/c ] = 2ac/((a + c) ) **

Harmonic mean of 4 numbers a, b, c, d = 4/((1/a + 1/b + 1/c +1/d) )

i.e. In an HP if you take three consecutive terms, middle term will be the harmonic mean of the first term and the third term.

**Application of Harmonic Mean**

Harmonic mean helps to find out the **average speed of a journey**. The average speed when the distances covered are equal = HM of the speed.s

Let’s say a person P, covers a distance A to B at a speed of “a” kmph and a distance B to A at a speed of “b” kmph.

#### As the distance is constant, we can find the average speed using the following formula **Average speed= 2ab/(a+b) kmph.**

**Illustrations:**

**1. A person travels from A to B with a speed of 60 km/hr and returns from B to A at 40 km/hr. What is the average speed for the whole journey?**

**Solution:**

Here since the distances covered are equal, average speed for the whole journey = Harmonic mean of the speeds =

HM of 60,40 = 2/[(1/60) + (1/40) ] = (2×60×40)/(60+40)= 48 km/hr.

**2. Amit covers first 1/3rd of the distance walking at 2 km/hr, second 1/3rd of the distance running at 3km/hr and the rest cycling at 6 km/hr. Find the average speed for the whole journey?**

**a) 3 km/hr **

**b) 3.66 km/hr **

**c) 4 km/hr **

**d) 5 km/hr.**

**Solution:**

Distances covered are equal for all 3 cases,

i.e. each time= 1/3rd of the total distance.

Hence, average speed = Harmonic mean of 2, 3 and 6

= 3/[(1/2)+ (1/3)+ (1/6) ] = 3 km/hr.

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