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HARMONIC PROGRESSION
If inverse of a sequence follows rule of an A.P. then it is said to be in harmonic
progression.
e.g. 1,1/2, 1/3, 1/4, 1/5 ...............
1/10, 1/7, 1/4, 1, – 1/2, ...........
In general
1/a, 1/a+d, 1/a+2d, ..................
Note:
Three convenient numbers in H.P. are
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
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,
or,
or,
∴
1/a, 1/H, 1/b are in A.P.
1/H – 1/a = 1/b – 1/H.
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.
Illustration:
Find the 4th and 8th term of the series 6, 4, 3, ……
Solution:
Consider1/6, /14, 1/3, ...... ∞
Here T2 – T1 = T3 – T2 = 1/12 ⇒ 1/6, 1/4, 1/3 is an A.P.
4th term of this A.P. = 1/6 + 3 × 1/12 = 1/6 + 1/4 = 5/12,
And the 8th term = 1/6 + 7 × 1/12 = 9/12.
Hence the 8th term of the H.P. = 12/9 = 4/3 and the 4th term = 12/5.
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:
Given that a, b, c are in H.P.
⇒ 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.
Some Important Results
• 1 + 2 + 3 +…+ n = n/2(n + 1) (sum of first n natural numbers).
• 12 + 22 + 32 +…+ n2 = n(n+1)(2n+1)/6 (sum of the squares of first n natural
numbers).
• 13 + 23 + 33 +…+ n3 = n2(n+1)2/4 = (1 + 2 + 3 +…+ n)2 (sum of the cubes of first n
natural numbers).
• (1 – x)–1 = 1 + x + x2 + x3 +…
–1 < x < 1.
–2
2
• (1 – x) = 1 + 2x + 3x +…
–1 < x < 1.
Illustration:
Find the nth term and the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 +…
Solution:
rth term of the series
= r(r+1).(r+3)=r3 + 4r2 + 3r
So sum of n terms
= Σnr=1 r3 + 4Σnr=1 r2 + 3Σnr=1 r
= (n(n+1)/2)2 + 4 n(n+1)(2n+1)/6 + 3n(n+1)/2 = n(n+1)/12 {3n 2 + 19n + 26}.
Illustration:
Find the sum of the series 1.n + 2(n–1) + 3.(n–2) +…+ n.1.
Solution:
The rth term of the series is
tr = (1 + (r – 1).1)(n + (r–1)(–1))
= r(n – r + 1) = r(n + 1) – r2
⇒ Sn= Σnr=1 trΣnr=1 (n+1)r – Σnr=1 r2 = (n+1) n.(n+1)/2 – n(n+1)(2n+1)/6
= n.(n+1)/2 [n + 1 – 2n+1/3] = n(n+1)(n–2)/6.