What is harmonic progression? A sequence is said to be in a harmonic progression if the reciprocal of its terms forms an arithmetic progression, i.e., the sequence a₁, a₂, a₃, a₄, a₅,…., a_n is a harmonic progression if the sequence 1/a₁, 1/a₂, 1/a₃, 1/a₄, 1/a₅,…., 1/a_n is an arithmetic progression or vice versa.

Harmonic Progression Formula

Below are a few of the important formulas and relationships used while dealing with harmonic progression problems.

1. Let AM is the arithmetic mean, GM is the geometric mean, and HM is the harmonic mean, then HM x AM = (GM)²

2. n^th term of HP = 1/(n^th term of AP)

3. Relationship between AM, GM and HM: AM > GM > HM

n^th Term of a Harmonic Progression

If a₁ and a₂ are the 1^st and 2^nd term of a harmonic progression, then the n^th term of an HP is given by

\begin{array}{l} T_{n} = \frac{1}{a + (n - 1) d} \end{array}

Since, a = 1/a₁

And, the common difference is

\begin{array}{l} d = \frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{a_{1} - a_{2}}{a_{1} a_{2}} \end{array}

Therefore,

\begin{array}{l} T_{n} = \frac{1}{\frac{1}{a_{1}} + (n - 1) \frac{a_{1} - a_{2}}{a_{1} a_{2}}} = \frac{a_{1} a_{2}}{a_{2} + (n - 1) (a_{1} - a_{2})} \end{array}

The n^th term of a harmonic progression from the end is given by:

In this case, the 1^st term = a_n and

d = – (Second Term – First Term) (Since the sequence is reversed)

\begin{array}{l} d = - [\frac{1}{a_{2}} - \frac{1}{a_{1}}] = - [\frac{a_{1} - a_{2}}{a_{1} a_{2}}] \end{array}

Therefore,

\begin{array}{l} T_{n} = \frac{1}{a_{n} + (n - 1) d} \end{array}

i.e.

\begin{array}{l} T_{n} = \frac{1}{\frac{1}{a_{n}} - (n - 1) \frac{a_{1} - a_{2}}{a_{1} a_{2}}} = \frac{a_{n} a_{1} a_{2}}{a_{1} a_{2} - a_{n} (n - 1) (a_{1} - a_{2})} \end{array}

Special Case:

1. If a, b, and c are in HP, then,

b = 2ac/(a + c)

And,

\begin{array}{l} \frac{a}{c} = \frac{a - b}{b - c} \end{array}

2. Three terms in a harmonic progression can be assumed as

\begin{array}{l} \frac{1}{a - d}, \frac{1}{a}, \frac{1}{a + d} \end{array}

Similarly, four and five convenient terms in HP can be taken as

\begin{array}{l} \frac{1}{a - 3 d}, \frac{1}{a - d}, \frac{1}{a + d}, \frac{1}{a + 3 d} \end{array}

and

\begin{array}{l} \frac{1}{a - 3 d}, \frac{1}{a - d}, \frac{1}{a}, \frac{1}{a + d}, \frac{1}{a + 3 d}, \end{array}

respectively.

Related Articles:

Harmonic Mean

Arithmetic Progression for IIT JEE

Relation between Arithmetic Mean, Geometric Mean and Harmonic Mean

If the two numbers are ‘a’ and ‘b’, a, b > 0, then,

Arithmetic Mean [AM] = (a+b)/2

Geometric Mean [GM] =√(ab)

Harmonic Mean [HM] = 2ab/(a + b)

∴ AM × HM = [(a + b)/2] × [2ab/(a + b)] = ab = √(ab)² = (GM)²

The means are in geometric progression.

So AM, GM, and HM follow G.P.

i.e.,

\begin{array}{l} G . M = \sqrt{A . M \times H . M} \end{array}

The difference between AM and GM is as follows:

AM – GM = [(a + b)/2] – √(ab)

\begin{array}{l} = \frac{a + b - 2 \sqrt{a b}}{2} \end{array}

\begin{array}{l} = \frac{(\sqrt{a} - \sqrt{b})^{2}}{2} \end{array}

i.e., AM > GM

Similarly,

G.M. – H.M. = √(ab) – (2ab)/(a+b)

= √(ab) (√a – √b)² /(a+b) > 0

So. GM > HM

Combining both results, AM > GM > HM.

Harmonic Progression Examples

Example 1: If the (p + 1)^th, (q + 1)^th, and (r + 1)^th terms of an arithmetic progression also form a geometric progression and the terms p, q, and r are in harmonic progression, then determine the ratio of the common difference to the 1st term of the arithmetic sequence.

Solution:

Let the 1^st term of the AP = a, and the common difference = d

Now, (p + 1)^th term of an AP = a + pd,

(q + 1)^th term of an AP = a + qd

And, (r + 1)^th term of an AP = a + rd

Since, (p + 1)^th, (q + 1)^th, and (r + 1)^th terms are in geometric progression

Therefore,

\begin{array}{l} (a + q d)^{2} = (a + p d) (a + r d) \end{array}

i.e.,

\begin{array}{l} a^{2} + (q d)^{2} + 2 a q d = a^{2} + a r d + a d p + r p d^{2} \end{array}

Or,

\begin{array}{l} q^{2} d + 2 a q = a r + a p + r p d \end{array}

Or,

\begin{array}{l} d (q^{2} - r p) = a (r + p - 2 q) \end{array}

Therefore,

\begin{array}{l} \frac{d}{a} = \frac{r + p - 2 q}{q^{2} - r p} \dots . (1) \end{array}

Since p, q, and r are in harmonic progression,

\begin{array}{l} \frac{2}{q} = \frac{1}{p} + \frac{1}{r} = \frac{p + r}{p r} \dots . (2) \end{array}

Now, from equation (1),

\begin{array}{l} \frac{d}{a} = \frac{r + p - 2 q}{q^{2} - r p} = \frac{2 (\frac{r + p}{2} - q)}{q (q - \frac{r p}{q})} \end{array}

Substituting the values of equation (2) in the above equation, we get,

\begin{array}{l} \frac{d}{a} = \frac{2 (\frac{pr}{q} - q)}{q (q - \frac{r p}{q})} = \frac{- 2}{q} \end{array}

Therefore, the ratio of the common difference to the 1^st term of the given arithmetic sequence is -2/q.

Example 2: If p², q²and ² are in arithmetic progression, then prove that q + r, r, r + p, and p + q are in harmonic progression.

Solution:

Given p², q², and r²are in arithmetic progression.

Therefore, p² + pq + qr + pr, q² + pq + qr + pr, and r²+ pq + qr + pr will also be in arithmetic progression.

Or, p (p + q) + r (q + p), q (q + p) + r (q + p), and r (r + p) + q (p + r)

Or, (p + q) (p + r), (q + r) (q + p), and (r + p) (r + q) will also in AP.

Now, on dividing each term by (p + q) (q + r) (r + p), we will get,

\begin{array}{l} \frac{1}{q + r}, \frac{1}{r + p}, and, \frac{1}{p + q} \end{array}

\begin{array}{l} Since \frac{1}{q + r}, \frac{1}{r + p}, and \frac{1}{p + q} are in AP . \end{array}

Therefore, q + r, r + p, and p + q are in harmonic progression.

Example 3: If a, b, and c are in harmonic progression, then find the value of

$\begin{array}{l} (\frac{1}{a} + \frac{1}{b} - \frac{1}{c}) (\frac{1}{b} + \frac{1}{c} - \frac{1}{a}) . \end{array}$

Solution:

Given: a, b, and c are in harmonic progression.

Using the formula for harmonic progression, we have,

\begin{array}{l} \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \dots . (1) \end{array}

Now,

\begin{array}{l} (\frac{1}{a} + \frac{1}{b} - \frac{1}{c}) (\frac{1}{b} + \frac{1}{c} - \frac{1}{a}) \end{array}

can be rewritten as:

\begin{array}{l} [\frac{1}{b} + \frac{1}{a} - \frac{1}{c}] [\frac{1}{b} - (\frac{1}{a} - \frac{1}{c})] \end{array}

Or,

\begin{array}{l} {(\frac{1}{b})}^{2} - {(\frac{1}{a} - \frac{1}{c})}^{2} \end{array}

Or,

\begin{array}{l} {(\frac{1}{b})}^{2} - [{(\frac{1}{a} + \frac{1}{c})}^{2} - \frac{4}{ac}] \end{array}

Now, substituting the values of equation (1) in the above equation, we get

\begin{array}{l} {(\frac{1}{b})}^{2} - [{(\frac{2}{b})}^{2} - \frac{4}{ac}] \end{array}

\begin{array}{l} = \frac{4}{ac} - \frac{3}{b^{2}} \end{array}

Therefore, the value of

\begin{array}{l} (\frac{1}{a} + \frac{1}{b} - \frac{1}{c}) (\frac{1}{b} + \frac{1}{c} - \frac{1}{a}) \end{array}

\begin{array}{l} = \frac{4}{ac} - \frac{3}{b^{2}} \end{array}

Example 4: If the p^th term of the harmonic progression is q, and the q^th term is p, find its (p + q)^th term.

Solution:

Given:

\begin{array}{l} T_{p} = q = \frac{1}{a + (p - 1) d} \end{array}

Or, a + (p – 1)d = 1/q ….(1)

And,

\begin{array}{l} T_{q} = p = \frac{1}{a + (q - 1) d} \end{array}

Or, a + (q – 1)d = 1/p ….(2)

Now, Equation (1) – Equation (2)

(p – 1) d – (q – 1) d = (1/q) – (1/p)

Or, d (p – q) = (p – q)/pq

Therefore, d = 1/pq

Substituting the values of d in equation (1), we get

\begin{array}{l} a + (p - 1) \frac{1}{p q} = \frac{1}{q} \end{array}

Or,

\begin{array}{l} a + \frac{1}{q} - \frac{1}{p q} = \frac{1}{q} \end{array}

Therefore, a = 1/pq

Now,

\begin{array}{l} T_{p + q} = \frac{1}{a + (n - 1) d} = \frac{1}{\frac{1}{p q} + (p + q - 1) \times \frac{1}{p q}} \end{array}

Therefore,

\begin{array}{l} T_{p + q} = \frac{p q}{p + q} \end{array}

Example 5: Let a₁, a₂, a₃,…., a_nbe in harmonic progression with a₁ = 5 and a₂₀ = 25. Find the least positive integer for which a_n < 0.

Solution:

Since the terms a₁, a₂, a₃, . . . . . . a_n are in harmonic progression. Therefore, the terms 1/a₁, 1/a₂, 1/a₃, . . . . . ., 1/a_n will be in arithmetic progression.

Given the first term of HP = a₁ = 5,

Therefore, the first term of AP = 1/a₁ = 1/5

Similarly, the 20th term of AP = 1/a₂₀ = 1/25

i.e., T₂₀ = 1/25 = (1/5) + 19d

Therefore, d = -4/475

Now, the least positive integer ‘n’ for which a_n < 0

i.e., a + (n – 1) d < 0

Or,

\begin{array}{l} \frac{1}{5} + (n - 1) \times \frac{- 4}{475} < 0 \end{array}

Or, (n – 1) > 95/4

Or, n > 24.75

Therefore, the least positive integer for which a_n < 0 is 24.75.

Example 6: Which number should be added to the numbers 13, 15, and 19 so that the resulting numbers are the consecutive terms of an H.P.?

Solution:

Suppose that x is to be added, then numbers 13, 15, 19 so that new numbers x + 13, 15 + x, 19 + x will be in H.P.

\begin{array}{l} \Rightarrow (15 + x) = \frac{2 (x + 13) (19 + x)}{x + 13 + x + 19} \\ \Rightarrow x^{2} + 31 x + 240 = x^{2} + 32 x + 247 \\ \Rightarrow x = - 7 \end{array}

Example 7: If x, y, z are in H.P., then find the value of expression log (x + z) + log (x − 2y + z).

Solution:

If x, y,z are in H.P., then

\begin{array}{l} y = \frac{2 x z}{x + z} \end{array}

Now,

\begin{array}{l} \log_{e} (x + z) + \log_{e} (x - 2 y + z) \\ \sum_{i = 1}^{n} = 3 \sum_{i = 1}^{n} i - 2 \sum_{i = 1}^{n} 1 = 3 \frac{n (n + 1)}{2} - 2 n \\ = \frac{n (3 n - 1)}{2} \\ = \log_{e} [(x + z) (x + z - \frac{4 x z}{x + z})] \\ = \log_{e} [{(x + z)}^{2} - 4 x z] \\ = \log_{e} {(x - z)}^{2} \\ = 2 \log_{e} (x - z) \end{array}

Harmonic Progression Formula

n^th Term of a Harmonic Progression

Relation between Arithmetic Mean, Geometric Mean and Harmonic Mean

Harmonic Progression Examples

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Harmonic Progression Formula

nth Term of a Harmonic Progression

Relation between Arithmetic Mean, Geometric Mean and Harmonic Mean

Harmonic Progression Examples

Comments

Leave a Comment Cancel reply

n^th Term of a Harmonic Progression