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) = G2

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

1. In a H.P 4/3, 3/2, 12/7, . ., find the 4^th and 7^th terms
2. 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

1. 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.


2. 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