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 = \(\mathbf{\frac{1}{\frac{1}{a_{1}}\;+\;(n\;-\;1)\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}}=\frac{a_{1}\;a_{2}}{a_{2}\;+\;(n\;-\;1)\;(a_{1}\;-\;a_{2})}}
\)

2. 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= \(\mathbf{\frac{1}{\frac{1}{a_{n}}\;-\;(n\;-\;1)\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}}=\frac{a_{n}\;a_{1}\;a_{2}}{a_{1}\;a_{2}\;-\;a_{n}\;(n\;-\;1)\;(a_{1}\;-\;a_{2})}}
\)

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

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

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

4. 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: T_p = q = \(\mathbf{\frac{1}{a\;+\;(p\;-\;1)\;d}}
\)

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

And, T_q= 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.