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Harmonic mean between two numbers a and b
Let H be the harmonic mean between two and number a and b.
So, a, H, b are in H.P.
or, 1/a, 1/H, 1/b are in A.P.
or, 1/H – 1/a = 1/b – 1/H.
or, 2/H = 1/a + 1/b = a+b/ab
∴ H =2ab/a+b
Similarly, we can find two harmonic mean between two number.
Let H1 and H2 be two harmonic mean between a and b.
So, a, H1, H2, b are in H.P.
or, 1/a, 1/H1 • 1/H2, 1/b are in A.P.
Using the formula,
tn = a + (n–1)d, we get,
1/b – 1/a = 3d., where ‘d’ is the common difference of A.P.
Or, 3d = a–b/ab
∴ d = a–b/3ab
So, 1/H1 = 1/a + d = 1/a + a–b/3ab = a+2b/3ab
and 1/H2 = 1/a + 2d = 1/a + 2(a–b)/3ab = 2a+2b/3ab
Summary of Important Notes
• If a and b are two non-zero numbers, then the harmonic mean of a and b is a number H such that the numbers a, H, b are in H.P. We have H = 1/H = 1/2 (1/a + 1/b) ⇒ H = 2ab/a+b.
• If a1, a2, ……, an are n non-zero numbers. then the harmonic mean H of these number is given by 1/H = 1/n (1/a1 + 1/a2 +...+ 1/an).
• 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.
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