 # Harmonic Progression IIT JEE Study Material

What is Harmonic Progression? A sequence is said to be in a harmonic progression if the reciprocal of its terms form an arithmetic progression i.e. the sequence

$$\begin{array}{l}a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, . . . . . . . . .,a_{n}\end{array}$$
is a harmonic progression if the sequence

$$\begin{array}{l}\mathbf{\frac{1}{a_{1}},\;\frac{1}{a_{2}},\;\frac{1}{a_{3}}\;, \;.\; .\; .\; ,\; \frac{1}{a_{n}}} \end{array}$$
is an arithmetic progression or vice versa.

## Harmonic Progression Formula

Below are few of the important formulas and relationships used while dealing with Harmonic progression problems:

1. Let AM is Arithmetic mean, GM is geometric mean and HM is 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

$$\begin{array}{l}a_{1} \end{array}$$
and
$$\begin{array}{l}a_{2} \end{array}$$
are the 1st and 2nd term of a harmonic progression then the nth term of a HP is given by:

Tn =

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

Since, a =

$$\begin{array}{l}\mathbf{\frac{1}{a_{1}}} \end{array}$$

And, the common difference (d) =

$$\begin{array}{l}\mathbf{\frac{1}{{a_{2}}}\;-\;\frac{1}{{a_{1}}}\;=\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}} \end{array}$$

Therefore, Tn =

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

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

In this case the 1st term = an and

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

i.e. d =

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

Therefore, Tn =

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

i.e. T=

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

Special Case:

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

$$\begin{array}{l}\mathbf{a_{2}\;=\;\frac{2\;a_{1}\;a_{3}}{a_{1}\;+\;a_{3}}} \end{array}$$

And,

$$\begin{array}{l}\mathbf{\frac{a_{1}}{a_{3}}\;=\;\frac{a_{1}\;-\;a_{2}}{a_{2}\;-\;a_{3}}} \end{array}$$

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] =

$$\begin{array}{l}\frac{2ab}{a+b}\end{array}$$

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

The means are in Geometric progression.

So AM, GM, HM follows G.P.

i.e. G.M. =

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

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 forms 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,

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

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

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

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

$$\begin{array}{l}\mathbf{\frac{-\;2}{q}} \end{array}$$

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 rare in Arithmetic Progression

Therefore,

$$\begin{array}{l}p^{2} + pq + qr + pr, q^{2} + pq + qr + pr \end{array}$$
, 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,

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

Since,

$$\begin{array}{l}\mathbf{\frac{1}{q\;+\;r},\;\frac{1}{r\;+\;p},\;and, \;\frac{1}{p\;+\;q}} \end{array}$$
are in AP, therefore,
$$\begin{array}{l}\mathbf{q\;+\;r,\;r\;+\;p,\;and, \;p\;+\;q} \end{array}$$
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 formula for harmonic progression, we have,

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

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:

i.e.

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

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 pth term of the harmonic progression is q, and the qth term is p, find its (p + q)th term.

Solution:

Given: Tp = q =

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

Or,

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

And, Tq= p =

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

Or,

$$\begin{array}{l}\mathbf{a\;+\;(q\;-\;1)\;d\;=\;\frac{1}{p}} \end{array}$$
. . . . . . . . . . . (2)

Now Equation (1) – Equation (2)

(p – 1) d – (q – 1) d =

$$\begin{array}{l}\mathbf{\frac{1}{q}\;-\;\frac{1}{p}} \end{array}$$

Or, d (p – q) =

$$\begin{array}{l}\mathbf{\frac{p\;-\;q}{p\;q}} \end{array}$$

Therefore, d =

$$\begin{array}{l}\mathbf{\frac{1}{p\;q}} \end{array}$$

Substituting the values of d in Equation (1) we get

i.e.

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

Or,

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

Therefore, a =

$$\begin{array}{l}\mathbf{\frac{1}{p\;q}}\end{array}$$

Now,

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

Therefore,

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

Example 5: Let

$$\begin{array}{l}a_{1}, a_{2}, a_{3}, . . . . . . a_{n}\end{array}$$
be in harmonic progression with
$$\begin{array}{l}a_{1}\end{array}$$
= 5 and
$$\begin{array}{l}a_{20}\end{array}$$
= 25. Find the least positive integer for which
$$\begin{array}{l}a_{n}\end{array}$$
< 0.

Solution:

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

$$\begin{array}{l}\mathbf{\frac{1}{a_{1}},\;\frac{1}{a_{2}},\;\frac{1}{a_{3}}\;.\;.\;.\;.\;\frac{1}{a_{n}}}\end{array}$$
will be in arithmetic progression.

Given, the first term of HP =

$$\begin{array}{l}a_{1}\end{array}$$
= 5,

Therefore, the first term of AP =

$$\begin{array}{l}\mathbf{\frac{1}{a_{1}}\;=\;\frac{1}{5}}\end{array}$$

Similarly, the 20th term of AP =

$$\begin{array}{l}\mathbf{\frac{1}{a_{20}}\;=\;\frac{1}{25}}\end{array}$$

I.e.

$$\begin{array}{l}T_{20}\end{array}$$
=
$$\begin{array}{l}\mathbf{\frac{1}{25}\;=\;\frac{1}{5}\;+\;19\;d}\end{array}$$

Therefore,

$$\begin{array}{l}\mathbf{d\;=\;-\;\frac{4}{475}}\end{array}$$

Now, the least positive integer ‘n’ for which

$$\begin{array}{l}a_{n}\end{array}$$
< 0

i.e. a + (n – 1) d < 0

Or,

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

Or,

$$\begin{array}{l}\mathbf{(n\;-\;1)\;>\;\frac{95}{4}}\end{array}$$

Or, n > 24.75

Therefore, the least positive integer for which

$$\begin{array}{l}a_{n}\end{array}$$
< 0 is 24.75.

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

Solution:

Suppose that x 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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