Harmonic progressions
Consider the sequence 1/3, 1/6, 1/9, 1/12, . . .
Let’s write the reciprocals of these terms. They are 3, 6, 9, 12, respectively, and are in arithmetic progression.
Such a progression is called a harmonic progression.
A progression is said to be a harmonic progression if the reciprocals of the terms form an arithmetic progression.
The n^{th} term of a harmonic progression
Examples
We know that a, a + d, a + 2d, . . . are in A.P.
The n^{th} term of this A.P. is a + (n  1)d.
Its reciprocal is 1/a + (n  1)d
So the n^{th} term of a H.P. is 1/a + (n  1)d
where the first term is 1/a and d is a constant independent of n.
The n^{th} term of an harmonic progression is 1/a + (n  1)d.
There is no concise general formula for the sum of n terms in an H.P.
Harmonic means
If a, b, c are in H.P, b is called the harmonic mean of a and c.
Now let us find the relationship.
a, b, c are in an H.P.
Therefore, 1/a, 1/b, 1/c are in A.P.
Therefore, 2/b = 1/a + 1/c = c+a/ac,
i.e., b = 2ac/(c + a)
The harmonic mean of a and c is 2ac/(c + a)
If a and b are positive integers, their arithmetic mean, geometric mean and harmonic mean are in a geometric progression.
Let the arithmetic mean, geometric mean and harmonic mean between a and b be A, G, H, respectively. Then
A = 

G = square root ab; H = 2ab/(a + b) 
A . H = 

(2ab/(a + b)) = ab = (square root ab) = G^{2} 
Hence, A, G, H are in a geometric progression.
Example 1
In a harmonic progression, the fourth term is 1/9 and the 13^{th} term is 1/27.
Write the harmonic progression.
Solution:
If the harmonic progression is
1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d), . . .
The 4^{th} term = 1/(a + 3d) =1/9. Therefore a + 3d = 9 —— (1)
The 13^{th} term = 1/(a + 12d) = 1/27 therefore
a + 12d = 27——— (2)
Solving (1) and (2) we get a = 3 and d = 2
So the harmonic progression is 1/3, 1/5, 1/7, 1/9, . . .
Example 2
Insert 4 harmonic means between 1/12 and 1/42
Solution:
Let the harmonic progression be
1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d), 1/(a + 4d), 1/(a + 5d)
Equating the first and last terms 1/a = 1/12 and
1/(a + 5d) = 1/42
Therefore,
a = 12 and a + 5d = 42, 5d = 42 – 12 = 30, d = 6.
Therefore the harmonic means are 1/18, 1/24, 1/30, 1/36.
Try these questions
Find the following
 In a H.P 4/3, 3/2, 12/7, . ., find the 4^{th} and 7^{th} terms
 In a HP the 3^{rd} term is 1/7 and the 7^{th} term is 1/5, then show that the 15^{th} term is 1
Answers
Find the following
 In a H.P 4/3, 3/2, 12/7, . . .,
Solution:
find the 4^{th} and 7^{th} terms
Solution:
Given H.P is 4/3, 3/2, 12/7, . . .
Therefore 3/4, 2/3, 7/12, . . . are in A.P.
a = 3/4, d = 2/3 – 3/4 = – 1/12
General term in A.P. = Tn = a + (n – 1)d
4^{th} term in A.P. = t4 = 3/4 + (4 – 1)(–1/12)
= 3/4 + 3(–1/12) = (3 – 1)/4 = 1/2
Therefore, 4^{th} item in H.P is 2
7^{th} term in A.P. = t7 = 3/4 + (7– 1)(–1/12)
= 3/4 – 1/2 = 1/4
Therefore, 7^{th} item in H.P is 4.
 In a HP the 3^{rd} term is 1/7 and the 7^{th} term is 1/5, then show that the 15^{th} term is 1
Solution:
The 3^{rd} and 7^{th} terms of a HP are 1/7 and 1/5 respectively.
Then 3^{rd} and 7^{th} terms of the corresponding A.P. are 7 and 5
Therefore, a + 2d = 7 ——— (1)
a + 6d = 5 ——— (2)
On solving equation 1 and 2, we get
d = –1/2 and a = 8
Therefore, 15^{th} term in A.P.
T15 = a + 14d
= 8 + 14(–1/2) = 8 – 7 = 1
Therefore 15^{th} term in H.P is 1