Let a, b and c form an H.P. Then 1/a, 1/b and 1/c form an A.P.
If a, b and c are in H.P. then 2/b = 1/a + 1/c, which can be simplified as b = 2ac/(a+c)
If ‘a’ and ‘b’ are two non-zero numbers then the sequence a, H, b is a H.P.
The n numbers H1, H2, ……,Hn are said to be harmonic means between a and b, if a, H1, H2 ……, Hn, b are in H.P. i.e. if 1/a, 1/H1, 1/H2, ..., 1/Hn, 1/b are in A.P. Let d be the common difference of the A.P., Then 1/b = 1/a + (n+1) d ⇒ d = a–b/(n+1)ab. Thus 1/H1 = 1/a + a–b/(n+1)ab, 1/H2 = 1/a + 2(a–n)/(n+1)ab, ……….. ………. 1/Hn = 1/a + n(a–b)/(n+1)ab.
If x1, x2, … xn are n non-zero numbers, then the harmonic mean ‘H’ of these numbers is given by 1/H = 1/n (1/x1 + 1/ x2 + ……. +1/ xn)
As the nth term of an A.P is given by an = a + (n-1)d, So the nth term of an H.P is given by 1/ [a + (n -1) d].
If we have a set of weights w1, w2, …. , wn associated with the set of values x1, x2, …. , xn, then the weighted harmonic mean is defined
Questions on Harmonic Progression are generally solved by first converting them into those of Arithmetic Progression.
If ‘a’ and ‘b’ are two positive real numbers then A.M x H.M = G.
The relation between the three means is defined as A.M > G.M > H.M
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
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