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,…., an is a harmonic progression if the sequence 1/a1, 1/a2, 1/a3, 1/a4, 1/a5,…., 1/an 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. nth term of HP = 1/(nth term of AP)

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

nth Term of a Harmonic Progression

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

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

Since, a = 1/a1

And, the common difference  is

\(\begin{array}{l}d =\mathbf{\frac{1}{{a_{2}}}\;-\;\frac{1}{{a_{1}}}\;=\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}}
\end{array} \)

Therefore,  

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

The nth term of a harmonic progression from the end is given by:

In this case, the 1st term = an and

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

\(\begin{array}{l}\mathbf{d = -\;\left [ \frac{1}{{a_{2}}}\;-\;\frac{1}{{a_{1}}} \right ]\;=\;-\;\left [ \frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}} \right ]}\end{array} \)

Therefore,  

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

i.e.  

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

Special Case:

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

b = 2ac/(a + c)

And,

\(\begin{array}{l}\mathbf{\frac{a}{c}\;=\;\frac{a\;-\;b}{b\;-\;c}}\end{array} \)

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

\(\begin{array}{l}\mathbf{\frac{1}{a\;-\;d},\;\frac{1}{a},\;\frac{1}{a\;+\;d}}
\end{array} \)
.

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

\(\begin{array}{l}\mathbf{\frac{1}{a\;-\;3d},\;\frac{1}{a\;-\;d},\;\frac{1}{a\;+\;d},\;\frac{1}{a\;+\;3d}}
\end{array} \)
and
\(\begin{array}{l}\mathbf{\frac{1}{a\;-\;3d},\;\frac{1}{a\;-\;d},\;\frac{1}{a},\;\frac{1}{a\;+\;d},\;\frac{1}{a\;+\;3d}},
\end{array} \)
 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)

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

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 1st 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,

\(\begin{array}{l}(a + qd)^{2} = (a + pd) (a + rd)
\end{array} \)

i.e.,

\(\begin{array}{l}a^{2} + (qd)^{2} + 2aqd = a^{2} + ard + adp +rpd^{2}
\end{array} \)

Or,

\(\begin{array}{l}q^{2}d + 2aq = ar + ap +rpd
\end{array} \)

Or,

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

Therefore,

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

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

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

Now, from equation (1),

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

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

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

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

Example 2: If p2, q2 and 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 r2 are 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,

\(\begin{array}{l}\mathbf{\frac{1}{q\;+\;r},\;\frac{1}{r\;+\;p},\;and, \;\frac{1}{p\;+\;q}}\end{array} \)

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

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

\(\begin{array}{l}\mathbf{\left ( \frac{1}{a}\;+\;\frac{1}{b}\;-\;\frac{1}{c} \right )\;\left ( \frac{1}{b}\;+\;\frac{1}{c}\;-\;\frac{1}{a} \right )}.\end{array} \)

Solution:

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

Using the formula for harmonic progression, we have,

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

Now,

\(\begin{array}{l}\mathbf{\left ( \frac{1}{a}\;+\;\frac{1}{b}\;-\;\frac{1}{c} \right )\;\left ( \frac{1}{b}\;+\;\frac{1}{c}\;-\;\frac{1}{a} \right )}
\end{array} \)
can be rewritten as:

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

Or,

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

Or,

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

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

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

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

Therefore, the value of

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

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

Example 4: If the pth term of the harmonic progression is q, and the qth 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

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

Or,

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

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,…., an be in harmonic progression with a1 = 5 and a20 = 25. Find the least positive integer for which an < 0.   

Solution:

Since the terms a1, a2, a3, . . . . . . an are in harmonic progression. Therefore, the terms 1/a1, 1/a2, 1/a3, . . . . . ., 1/an 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 an < 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 an < 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.

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

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 

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

Now,

\(\begin{array}{l}{{\log }_{e}}(x+z)+{{\log }_{e}}(x-2y+z) \\\sum\limits_{i=1}^{n}{{}}=3\sum\limits_{i=1}^{n}{i}-2\sum\limits_{i=1}^{n}{1}=3\frac{n(n+1)}{2}-2n\\=\frac{n(3n-1)}{2} \\={{\log }_{e}}\left[ (x+z)\,\left( x+z-\frac{4xz}{x+z} \right) \right] \\={{\log }_{e}}[{{(x+z)}^{2}}-4xz]\\={{\log }_{e}}{{(x-z)}^{2}}\\=2{{\log }_{e}}(x-z)\end{array} \)
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