A series of non-zero numbers is said to be harmonic progression
(abbreviated H.P.) if the series obtained by taking reciprocals of the
corresponding terms of the given series is an arithmetic progression.
For example, the series 1 +1/4 +1/7 +1/10 +..... is an H.P. since the series
obtained by taking reciprocals of its corresponding terms i.e. 1 +4 +7 +10
+... is an A.P.
A general H.P. is 1/a + 1/(a + d) + 1(a + 2d) + ...
nth term of an H.P. = 1/[a +(n -1)d]
Three numbers a, b, c are in H.P. iff 1/a, 1/b, 1/c are in A.P.
i.e. iff 1/a + 1/c = 2/b
i.e. iff b= 2ac/(a + c)
Thus the H.M. between a and b is H = 2ac/(a + c)
The 7th term of an H.P. is 1/10 and 12th term is 1/25 Find the 20th term, and the nth term.
Let the H.P. be 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]
Prove that three quantities a, b, c are in A.P., G.P., or H.P. iff
(a-b)/(b-c) = a/a, a/b or a/c respectively.
a, b, c are in A.P. iff b -a = c -b i.e.
iff (a-b)/(b-c) = 1 = a/a
Also a, b, c are in G.P.
iff b/a = c/b i.e iff 1 - b/a = 1 - c/b
i.e. iff (a-b)/a = (b-c)/b
i.e. iff (a-b)/(b-c) = a/b
Similarly a, b, c are in H.P.
iff 1/b - 1/a = 1/c - 1/b
i.e. iff (a-b)/ab = (b-c)/bc
i.e. iff (a-b)/(b-c) = ab/bc = a/c
If A, G, H are arithmetic, geometric and harmonic means between two distinct, positive real numbers a and b, show that
Here A = A.M. between a and b = (a+b)/2,
G = G.M. between a and b =
(ab),
H = H.M. between a and b = 2ab/(a+b)
(i) A.H = [(a+b)/2].[2ab/(a+b)] = [(ab)]² = G²
(ii) A - G = (a+b)/2 - (ab) = (1/2)[a +b -2(ab)]
= (1/2)(a -b)²> 0
as (a -b)² is square of a non-zero
real number.
Also G - H = ab - 2ab/(a+b) = [(ab)/(a+b)](a
+ b - 2ab)
= [(ab)/(a+b)](a -b)²> 0
Hence A > G > H.