# Harmonic Progression IIT JEE Study Material

A sequence is said to be in a harmonic progression if the reciprocal of its terms form an arithmetic progression i.e. the sequence $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, . . . . . . . . a_{n}$ is a harmonic progression if the sequence $\mathbf{\frac{1}{a_{1}},\;\frac{1}{a_{2}},\;\frac{1}{a_{3}}\;, \;.\; .\; .\; ,\; \frac{1}{a_{n}}}$
is an arithmetic progression or vice versa.

1. If $a_{1}$
and $a_{2}$
are the 1st and 2nd term of a harmonic progression then the nth term of a HP is given by:

Tn = $\mathbf{\frac{1}{a\;+\;(n\;-\;1)\;d}}$

Since, a = $\mathbf{\frac{1}{a_{1}}}$

And, the common difference (d) = $\mathbf{\frac{1}{{a_{2}}}\;-\;\frac{1}{{a_{1}}}\;=\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}}$

Therefore, Tn = $\mathbf{\frac{1}{\frac{1}{a_{1}}\;+\;(n\;-\;1)\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}}=\frac{a_{1}\;a_{2}}{a_{2}\;+\;(n\;-\;1)\;(a_{1}\;-\;a_{2})}}$

1. 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 = $\mathbf{-\;\left [ \frac{1}{{a_{2}}}\;-\;\frac{1}{{a_{1}}} \right ]\;=\;-\;\left [ \frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}} \right ]}$

Therefore, Tn = $\mathbf{\frac{1}{a_{n}\;+\;(n\;-\;1)\;d}}$

i.e. Tn= $\mathbf{\frac{1}{\frac{1}{a_{n}}\;-\;(n\;-\;1)\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}}=\frac{a_{n}\;a_{1}\;a_{2}}{a_{1}\;a_{2}\;-\;a_{n}\;(n\;-\;1)\;(a_{1}\;-\;a_{2})}}$

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

$\mathbf{a_{2}\;=\;\frac{2\;a_{1}\;a_{3}}{a_{1}\;+\;a_{3}}}$

And, $\mathbf{\frac{a_{1}}{a_{3}}\;=\;\frac{a_{1}\;-\;a_{2}}{a_{2}\;-\;a_{3}}}$

1. Three terms in a harmonic progression can be assumed as $\mathbf{\frac{1}{a\;-\;d},\;\frac{1}{a},\;\frac{1}{a\;+\;d}}$
. Similarly four and five convenient terms in HP can be taken as $\mathbf{\frac{1}{a\;-\;3d},\;\frac{1}{a\;-\;d},\;\frac{1}{a\;+\;d},\;\frac{1}{a\;+\;3d}}$
and $\mathbf{\frac{1}{a\;-\;3d},\;\frac{1}{a\;-\;d},\;\frac{1}{a},\;\frac{1}{a\;+\;d},\;\frac{1}{a\;+\;3d}}$
respectively.

## Harmonic Progression IIT JEE Problems

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, $(a + qd)^{2} = (a + pd) (a + rd)$

i.e. $a^{2} + (qd)^{2} + 2aqd = a_{2} + ard + adp +rpd^{2}$

Or, $q^{2}d + 2aq = ar + ap +rpd$

Or, $d (q^{2} + – rp) = a (r + p – 2q)$

Therefore, $\mathbf{\frac{d}{a}\;=\;\frac{r\;+\;p\;-\;2q}{q^{2}\;-\;rp}\;.\;.\;.\;.\;.\;.\;.\;\;(1)}$

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

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

Now from Equation (1),

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

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

$\mathbf{\frac{d}{a}\;=\;\frac{2\left ( \frac{pr}{q} \;-\;q\right )}{q\;\left ( q\;-\;\frac{r\;p}{q} \right )}\;=\;\frac{-\;2}{q}}$

Therefore, the ratio of the common difference to the 1st term of the given arithmetic sequence is $\mathbf{\frac{-\;2}{q}}$

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, $p^{2} + pq + qr + pr, q^{2} + pq + qr + pr$
, 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,

$\mathbf{\frac{1}{q\;+\;r},\;\frac{1}{r\;+\;p},\;and, \;\frac{1}{p\;+\;q}}$

Since, $\mathbf{\frac{1}{q\;+\;r},\;\frac{1}{r\;+\;p},\;and, \;\frac{1}{p\;+\;q}}$
are in AP, therefore, $\mathbf{q\;+\;r,\;r\;+\;p,\;and, \;p\;+\;q}$
are in harmonic progression.

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

$\mathbf{\left ( \frac{1}{a}\;+\;\frac{1}{b}\;-\;\frac{1}{c} \right )\;\left ( \frac{1}{b}\;+\;\frac{1}{c}\;-\;\frac{1}{a} \right )}$

Solution:

Given: a, b, and c are in Harmonic Progression.

Therefore, $\mathbf{\frac{2}{b}\;=\;\frac{1}{a}\;+\;\frac{1}{c}}$
. . . . . . . . . (1)

Now, $\mathbf{\left ( \frac{1}{a}\;+\;\frac{1}{b}\;-\;\frac{1}{c} \right )\;\left ( \frac{1}{b}\;+\;\frac{1}{c}\;-\;\frac{1}{a} \right )}$
can be rewritten as:

i.e. $\mathbf{\left [ \frac{1}{b}\;+\;\frac{1}{a}\;-\;\frac{1}{c} \right ]\;\left [ \frac{1}{b}\;-\;\left ( \;\frac{1}{a}\;-\;\frac{1}{c} \right ) \right ]}$

Or, $\mathbf{\left (\frac{1}{b} \right )^{2}\;-\;\left ( \;\frac{1}{a}\;-\;\frac{1}{c} \right )^{2}}$

Or, $\mathbf{\left (\frac{1}{b} \right )^{2}\;-\;\left [ \left ( \;\frac{1}{a}\;+\;\frac{1}{c} \right )^{2}\;-\;\frac{4}{ac} \right ]}$

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

$\mathbf{\left (\frac{1}{b} \right )^{2}\;-\;\left [ \left ( \;\frac{2}{b} \;\right )^{2}\;-\;\frac{4}{ac} \right ]}$
= $\mathbf{\frac{4}{ac}\;-\;\frac{3}{b^{2}}}$

Therefore, the value of $\mathbf{\left ( \frac{1}{a}\;+\;\frac{1}{b}\;-\;\frac{1}{c} \right )\;\left ( \frac{1}{b}\;+\;\frac{1}{c}\;-\;\frac{1}{a} \right )}$
= $\mathbf{\frac{4}{ac}\;-\;\frac{3}{b^{2}}}$

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 = $\mathbf{\frac{1}{a\;+\;(p\;-\;1)\;d}}$

Or, $\mathbf{a\;+\;(p\;-\;1)\;d\;=\;\frac{1}{q}}$
. . . . . . . . . (1)

And, Tq= p = $\mathbf{\frac{1}{a\;+\;(q\;-\;1)\;d}}$

Or, $\mathbf{a\;+\;(q\;-\;1)\;d\;=\;\frac{1}{p}}$
. . . . . . . . . . . (2)

Now Equation (1) – Equation (2)

(p – 1) d – (q – 1) d = $\mathbf{\frac{1}{q}\;-\;\frac{1}{p}}$

Or, d (p – q) = $\mathbf{\frac{p\;-\;q}{p\;q}}$

Therefore, d = $\mathbf{\frac{1}{p\;q}}$

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

i.e. $\mathbf{a\;+\;(p\;-\;1)\;\frac{1}{p\;q}\;=\;\frac{1}{q}}$

Or, $\mathbf{a\;+\;\frac{1}{q}\;-\;\frac{1}{p\;q}\;=\;\frac{1}{q}}$

Therefore, a = $\mathbf{\frac{1}{p\;q}}$

Now, $T_{p+q}$ = $\mathbf{\frac{1}{a\;+\;(n\;-\;1)\;d}\;=\;\frac{1}{\frac{1}{p\;q}\;+\;(p\;+\;q\;-\;1)\;\times \;\frac{1}{p\;q}}}$

Therefore, $T_{p + q}$ = $\mathbf{\frac{p\;q}{p\;+\;q}}$

Example 5: Let $a_{1}, a_{2}, a_{3}, . . . . . . a_{n}$ be in harmonic progression with $a_{1}$ = 5 and $a_{20}$ = 25. Find the least positive integer for which $a_{n}$ < 0.

Solution:

Since the terms a1, a2, a3, . . . . . . an are in harmonic progression. Therefore, the terms $\mathbf{\frac{1}{a_{1}},\;\frac{1}{a_{2}},\;\frac{1}{a_{3}}\;.\;.\;.\;.\;\frac{1}{a_{n}}}$ will be in arithmetic progression.

Given, the first term of HP = $a_{1}$  = 5,

Therefore, the first term of AP = $\mathbf{\frac{1}{a_{1}}\;=\;\frac{1}{5}}$

Similarly, the 20th term of AP = $\mathbf{\frac{1}{a_{20}}\;=\;\frac{1}{25}}$

I.e. $T_{20}$ = $\mathbf{\frac{1}{25}\;=\;\frac{1}{5}\;+\;19\;d}$

Therefore, $\mathbf{d\;=\;-\;\frac{4}{475}}$

Now, the least positive integer ‘n’ for which $a_{n}$ < 0

i.e. a + (n – 1) d < 0

Or, $\mathbf{\frac{1}{5}\;+\;(n\;-\;1)\;\times \;\frac{-4}{475}\;<\;0}$

Or, $\mathbf{(n\;-\;1)\;>\;\frac{95}{4}}$

Or, n > 24.75

Therefore, the least positive integer for which $a_{n}$ < 0 is 24.75.