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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 a1,a2,a3,a4,a5,.........,ana_{1}, a_{2}, a_{3}, a_{4}, a_{5}, . . . . . . . . .,a_{n} is a harmonic progression if the sequence

1a1,  1a2,  1a3  ,  .  .  .  ,  1an\mathbf{\frac{1}{a_{1}},\;\frac{1}{a_{2}},\;\frac{1}{a_{3}}\;, \;.\; .\; .\; ,\; \frac{1}{a_{n}}} is an arithmetic progression or vice versa.

Harmonic Progression Formula

Below is 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 a1a_{1} and a2a_{2} are the 1st and 2nd term of a harmonic progression then the nth term of a HP is given by:

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

Since, a = 1a1\mathbf{\frac{1}{a_{1}}}

And, the common difference (d) = 1a2  −  1a1  =  a1  −  a2a1  a2\mathbf{\frac{1}{{a_{2}}}\;-\;\frac{1}{{a_{1}}}\;=\;\frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}}}

Therefore, Tn = 11a1  +  (n  −  1)  a1  −  a2a1  a2=a1  a2a2  +  (n  −  1)  (a1  −  a2)\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})}}

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 = −  [1a2  −  1a1]  =  −  [a1  −  a2a1  a2]\mathbf{-\;\left [ \frac{1}{{a_{2}}}\;-\;\frac{1}{{a_{1}}} \right ]\;=\;-\;\left [ \frac{a_{1}\;-\;a_{2}}{a_{1}\;a_{2}} \right ]}

 

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

 

i.e. Tn= 11an  −  (n  −  1)  a1  −  a2a1  a2=an  a1  a2a1  a2  −  an  (n  −  1)  (a1  −  a2)\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})}}

 

Special Case:

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

a2  =  2  a1  a3a1  +  a3\mathbf{a_{2}\;=\;\frac{2\;a_{1}\;a_{3}}{a_{1}\;+\;a_{3}}}

And, a1a3  =  a1  −  a2a2  −  a3\mathbf{\frac{a_{1}}{a_{3}}\;=\;\frac{a_{1}\;-\;a_{2}}{a_{2}\;-\;a_{3}}}

 

2. Three terms in a harmonic progression can be assumed as 1a  −  d,  1a,  1a  +  d\mathbf{\frac{1}{a\;-\;d},\;\frac{1}{a},\;\frac{1}{a\;+\;d}} .

Similarly four and five convenient terms in HP can be taken as 1a  −  3d,  1a  −  d,  1a  +  d,  1a  +  3d\mathbf{\frac{1}{a\;-\;3d},\;\frac{1}{a\;-\;d},\;\frac{1}{a\;+\;d},\;\frac{1}{a\;+\;3d}} and 1a  −  3d,  1a  −  d,  1a,  1a  +  d,  1a  +  3d\mathbf{\frac{1}{a\;-\;3d},\;\frac{1}{a\;-\;d},\;\frac{1}{a},\;\frac{1}{a\;+\;d},\;\frac{1}{a\;+\;3d}} 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, HM follows G.P.

i.e. G.M. =√A.M. × H.M.

The difference between AM and GM is as follows:

AM – GM = a+b/2 – √ab

=(√a^2)+(√b)–2√a√b/2

i.e. AM > GM

Similarly,

G.M. – H.M. = √ab – 2ab/a+b

=√ab/a+b (√a – √b)2 > 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, (a+qd)2=(a+pd)(a+rd)(a + qd)^{2} = (a + pd) (a + rd)

i.e. a2+(qd)2+2aqd=a2+ard+adp+rpd2a^{2} + (qd)^{2} + 2aqd = a_{2} + ard + adp +rpd^{2}

Or, q2d+2aq=ar+ap+rpdq^{2}d + 2aq = ar + ap +rpd

Or, d(q2+–rp)=a(r+p–2q)d (q^{2} + – rp) = a (r + p – 2q)

Therefore, da  =  r  +  p  −  2qq2  −  rp  .  .  .  .  .  .  .    (1)\mathbf{\frac{d}{a}\;=\;\frac{r\;+\;p\;-\;2q}{q^{2}\;-\;rp}\;.\;.\;.\;.\;.\;.\;.\;\;(1)}

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

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

Now from Equation (1),

da  =  r  +  p  −  2qq2  −  r  p  =  2(r  +  p2  −  q)q  (q  −  r  pq)\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,

da  =  2(prq  −  q)q  (q  −  r  pq)  =  −  2q\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 −  2q\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, p2+pq+qr+pr,q2+pq+qr+prp^{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,

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

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

(1a  +  1b  −  1c)  (1b  +  1c  −  1a)\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.

Using formula for harmonic progression, we have, 2b  =  1a  +  1c\mathbf{\frac{2}{b}\;=\;\frac{1}{a}\;+\;\frac{1}{c}} . . . . . . . . . (1)

Now, (1a  +  1b  −  1c)  (1b  +  1c  −  1a)\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. [1b  +  1a  −  1c]  [1b  −  (  1a  −  1c)]\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, (1b)2  −  (  1a  −  1c)2\mathbf{\left (\frac{1}{b} \right )^{2}\;-\;\left ( \;\frac{1}{a}\;-\;\frac{1}{c} \right )^{2}}

Or, (1b)2  −  [(  1a  +  1c)2  −  4ac]\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:

(1b)2  −  [(  2b  )2  −  4ac]\mathbf{\left (\frac{1}{b} \right )^{2}\;-\;\left [ \left ( \;\frac{2}{b} \;\right )^{2}\;-\;\frac{4}{ac} \right ]} = 4ac  −  3b2\mathbf{\frac{4}{ac}\;-\;\frac{3}{b^{2}}}

Therefore, the value of (1a  +  1b  −  1c)  (1b  +  1c  −  1a)\mathbf{\left ( \frac{1}{a}\;+\;\frac{1}{b}\;-\;\frac{1}{c} \right )\;\left ( \frac{1}{b}\;+\;\frac{1}{c}\;-\;\frac{1}{a} \right )} = 4ac  −  3b2\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 = 1a  +  (p  −  1)  d\mathbf{\frac{1}{a\;+\;(p\;-\;1)\;d}}

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

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

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

Now Equation (1) – Equation (2)

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

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

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

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

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

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

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

Now, Tp+qT_{p+q} = 1a  +  (n  −  1)  d  =  11p  q  +  (p  +  q  −  1)  ×  1p  q\mathbf{\frac{1}{a\;+\;(n\;-\;1)\;d}\;=\;\frac{1}{\frac{1}{p\;q}\;+\;(p\;+\;q\;-\;1)\;\times \;\frac{1}{p\;q}}}

Therefore, Tp+qT_{p + q} = p  qp  +  q\mathbf{\frac{p\;q}{p\;+\;q}}

Example 5: Let a1,a2,a3,......ana_{1}, a_{2}, a_{3}, . . . . . . a_{n} be in harmonic progression with a1a_{1} = 5 and a20a_{20} = 25. Find the least positive integer for which ana_{n} < 0.   

Solution:

Since the terms a1, a2, a3, . . . . . . an are in harmonic progression. Therefore, the terms 1a1,  1a2,  1a3  .  .  .  .  1an\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 = a1a_{1}  = 5,

Therefore, the first term of AP = 1a1  =  15\mathbf{\frac{1}{a_{1}}\;=\;\frac{1}{5}}

Similarly, the 20th term of AP = 1a20  =  125\mathbf{\frac{1}{a_{20}}\;=\;\frac{1}{25}}

I.e. T20T_{20} = 125  =  15  +  19  d\mathbf{\frac{1}{25}\;=\;\frac{1}{5}\;+\;19\;d}

Therefore, d  =  −  4475\mathbf{d\;=\;-\;\frac{4}{475}}

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

i.e. a + (n – 1) d < 0

Or, 15  +  (n  −  1)  ×  −4475  <  0\mathbf{\frac{1}{5}\;+\;(n\;-\;1)\;\times \;\frac{-4}{475}\;<\;0}

Or, (n  −  1)  >  954\mathbf{(n\;-\;1)\;>\;\frac{95}{4}}

Or, n > 24.75

Therefore, the least positive integer for which ana_{n} < 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.

⇒(15+x)=2(x+13)(19+x)x+13+x+19⇒x2+31x+240=x2+32x+247⇒x=−7\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\\

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 y=2xzx+zy=\frac{2xz}{x+z}

Now, log⁡e(x+z)+log⁡e(x−2y+z)∑i=1n=3∑i=1ni−2∑i=1n1=3n(n+1)2−2n=n(3n−1)2=log⁡e[(x+z) (x+z−4xzx+z)]=log⁡e[(x+z)2−4xz]=log⁡e(x−z)2=2log⁡e(x−z){{\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)