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Harmonic progression is an important topic of sequence and series which is closely related to the arithmetic progression. We shall discuss arithmetic and geometric progressions in brief as they have been discussed in detail in the earlier sections.
Arithmetic Progression: A progression in which there is a fixed difference between every pair of two terms i.e. the progression of the form a, a+d, a+2d, ….. is known as an arithmetic progression.
Geometric Progression: The sequence or progression of the form a, ar, ar^{2}, …. is said to be a geometric progression.
The simplest way to define a harmonic progression is that if the inverse of a sequence follows the rule of an arithmetic progression then it is said to be in harmonic progression. This simply means that if a, a+d, a+2d, ….. is an A.P. then 1/a, 1/(a+d), 1/(a+2d), …… is an H.P.
For example, the series 1 +1/4 +1/7 +1/10 +..... is an example of harmonic progression, since the series obtained by taking reciprocals of its corresponding terms i.e. 1 +4 +7 +10 +... is an A.P.
Let a, b and c form an H.P. Then as discussed above, 1/a, 1/b and 1/c form an A.P. Then 2/b = 1/a + 1/c. Hence, b = 2ac/ (a+c). This ‘b’ is called as the harmonic mean of a and c.
Illustration: We have a H.P. whose 7^{th} term is 1/10 and 12^{th} term is 1/25. Find the 20th term and hence the nth term.
Solution: We know that the general form of H.P. is
1/a + 1/(a + d) + 1(a + 2d) + ... The 7th term = 1/(a + 6d) = 1/10 => a +6 d = 10 The 12th term = 1/(a + 11d) = 1/25 => a +11 d = 25 Solving these two equations, a = -8, d = 3 Hence 20th term = 1/(a+19d) = 1/[-8 + 9(3)] = 1/49 and nth term = 1/[a +(n -1)d] = 1/[-8 +(n -1) 3] = 1/[3n - 11]
Illustration: If a, b, c are in H.P., show that a/b+c, b/c+a, c/a+b are also in H.P.
Solution: It is given in the question that a, b and c are in H.P.
By definition it follows that 1/a, 1/b, 1/c are in A.P
⇒ (a+b+c)/a, (a+b+c)/b, (a+b+c)/c are in A.P.
⇒ 1+ (b+c)/a, 1 + (c+a)/b, 1 + (a+b)/c are in A.P
⇒ (b+c)/a, (c+a)/b, (a+b)/c are in A.P.
⇒ a/b+c, b/c+a, c/a+b are in H.P.
Illustration: If a^{2}, b^{2} and c^{2} are in A.P., then prove that b+c, c+a, a+b are in H.P.
Solution: It is given that a^{2}, b^{2} and c^{2} are in A.P.
Hence, adding (ab + bc + ca) to each term
So, a^{2} + ab + bc + ca, b^{2} + ab + bc + ca and c^{2} + ab + bc + ca will also be in A.P
Hence, a(a+b) + c(a+b), b(a+b) + c(a+b), c(c+b) + a(b+c) are in A.P
i.e. (a+b) (a+c), (a+b) (c+b), (a+c) (c+b) are in A.P
dividing each term by (a+b) (c+b) (a+c) we get,
1/b+c, 1/c+a, 1/a+b are in A.P
Hence, (b+c), (c+a), (a+b) are in H.P.
As the nth term of an A.P is given by a_{n} = a + (n-1)d,
So the nth term of an H.P is given by 1/ [a + (n -1) d].
Let a and b be two positive real numbers, then
A.M x H.M = G.M^{2}
The relation between the three types of means is
A.M > G.M > H.M.
The figure given below further illustrates this relation
If we need to find three numbers in a H.P. then they should be assumed as 1/a–d, 1/a, 1/a+d
Four convenient numbers in H.P. are
1/a–3d, 1/a–d, 1/a+d, 1/a+3d
Five convenient numbers in H.P. are
1/a–2d, 1/a–d, 1/a, 1/a+d, 1/a+2d
Majority of the questions of H.P. are solved by first converting them into A.P.
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