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,.........,an is a harmonic progression if the sequence
a11,a21,a31,...,an1 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 a1 and a2 are the 1st and 2nd term of a harmonic progression then the nth term of a HP is given by:
Tn = a+(n−1)d1
Since, a = a11
And, the common difference (d) = a21−a11=a1a2a1−a2
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] = a+b2ab
∴ 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)
i.e. AM > GM
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.
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
Since p, q, and, r are in harmonic progression, therefore,
q2=p1+r1=prp+r. . . . . . . . . (2)
Now from Equation (1),
Substituting the values of equation (2) in the above equation we get,
Therefore, the ratio of the common difference to the 1st term of the given arithmetic sequence is q−2
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.
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,
Since, q+r1,r+p1,and,p+q1 are in AP, 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
Given: a, b, and c are in Harmonic Progression.
Using formula for harmonic progression, we have, b2=a1+c1 . . . . . . . . . (1)
Now, (a1+b1−c1)(b1+c1−a1) can be rewritten as:
Now, substituting the values of equation (1) in the above equation we get:
(b1)2−[(b2)2−ac4] = ac4−b23
Therefore, the value of (a1+b1−c1)(b1+c1−a1) = ac4−b23
Example 4: If pth term of the harmonic progression is q, and the qth term is p, find its (p + q)th term.
Given: Tp = q = a+(p−1)d1
Or, a+(p−1)d=q1. . . . . . . . . (1)
And, Tq= p = a+(q−1)d1
Or, a+(q−1)d=p1. . . . . . . . . . . (2)
Now Equation (1) – Equation (2)
(p – 1) d – (q – 1) d = q1−p1
Or, d (p – q) = pqp−q
Therefore, d = pq1
Substituting the values of d in Equation (1) we get
Therefore, a = pq1
Now, Tp+q = a+(n−1)d1=pq1+(p+q−1)×pq11
Therefore, Tp+q = p+qpq
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.
Since the terms a1, a2, a3, . . . . . . an are in harmonic progression. Therefore, the terms a11,a21,a31....an1 will be in arithmetic progression.
Given, the first term of HP = a1 = 5,
Therefore, the first term of AP = a11=51
Similarly, the 20th term of AP = a201=251
I.e. T20 = 251=51+19d
Now, the least positive integer ‘n’ for which an < 0
i.e. a + (n – 1) d < 0
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, 19 so that the resulting numbers are the consecutive terms of a H.P.?
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.