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.
Solved Examples Based on Harmonic mean 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.
• (1 – x)–2 = 1 + 2x + 3x2 +…
–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 So sum of n terms
= =
r(r+1).(r+3)=r3 + 4r2 + 3r
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 {3n2 + 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 tr Σnr=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.
Relation between A.M., G.M. and H.M.
Let there are two numbers ‘a’ and ‘b’, a, b > 0 then AM = a+b/2
GM =√ab HM =2ab/a+b ∴
AM × HM =a+b/2 × 2ab/a+b = ab = (√ab)2 = (GM)2
Note that these means are in G.P. Hence AM.GM.HM follows the rules of G.P.
i.e. G.M. =√A.M. × H.M. Now, let us see the difference between AM and GM AM – GM =a+b/2 – √ab
=(√a2)+(√b)–2√a√b/2 i.e. AM > GM Similarly, G.M. – H.M. = √ab –2ab/a+b
=√ab/a+b (√a – √b)2 > 0 So. GM > HM Combining both results, we get AM > GM > HM
…….. (12)
All sequences of numbers cannot be put into A.P./G.P./H.P. Let us study these. Important Points:
r3 (r – 1)3 = 3 r2 – 3r + 1 r = 1 : 13 – 0 = 3 . 12 – 3 . 1 + 1 r = 2 : 23 – 13 = 3 . 22 – 3 . 2 + 1 r = 3 : 33 – 23 = 3 . 32 – 3 . 3 + 1 r = n : n3 – (n–1)3 = 3.(n2) – 3(n) + 1
Adding n3 = 3 (12 + 22 +…+ n2) –3 (1 + 2 + 3 +…+ n) + (1 + 1 +…+ n times) n3 = 3 Σnr=1 r2 – 3 (n(n+1))/2 + n ⇒
3 Σnr=1 r2 = n3 + 3n(n+1)/2 – n = n/2 (2n2 + 3n + 1)
⇒
Σnr=1 r2 = n(n+1)(2n+1)/6
Method of Differences
Suppose a1, a2, a3, …… is a sequence such that the sequence a2 – a1, a3 – a3, … is either an. A.P. or a G.P. The nth term, of this sequence is obtained as follows: S = a1 + a2 + a3 +…+ an–1 + an S = a1 + a2 +…+ an–2 + an–1 + an
…… (1) …… (2)
Subtracting (2) from (1), we get, a n = a1 + [(a2–a1) + (a3–a2)+…+(an–an–1)]. Since the terms within the brackets are either in an A.P. or a G.P., we can find the value of an, the n th term. We can now find the sum of the n terms of the sequence as S = Σnk=1 ak. F corresponding to the sequence a1, a2, a3, ……, an, there exists a sequence b0, b1, b2, ……, bn such that ak = bk – bk–1, then sum of n terms of the sequence a1, a2, ……, an is bn – b0. Illustration:
Find the sum of 1st n terms of the series 5, 7, 11, 17, 25, … Solution:
Let S = 5 + 7 + 11 + 17 + 25 +…+ tn. Also S = 5 + 7 + 11 + 17 +… + tn–1 + tn. Subtracting we get
th
0 = 5 + 2 + 4 + 6 + 8 +…+ n term – tn
tn = 5 + 2 {n(n–1)/2}. or tn = n2 – n + 5 2 ⇒ Sn = Σn – Σn + 5Σ 1 = n(n+1)(2n+1)/6 – n(n+1)/2 + 5n = n/6 {(n + 1)(2n + 1) –3(n + 1) + 30} = n/6 (2n2 + 28). ⇒
Illustrations based on Vn method Illustration:
Find the sum of the series 1/1.2.3.4 + 1/2.3.4.5 +...... n terms. Solution:
Let Tr =1/r(r+1)(r+2)(r+3) = 1/3 – [(r+3)–r]/r(r+1)(r+2)(r+3) =1/3 [1/r(r+1)(r+2) – 1/(r+1)(r+2)(r+3)] = 1/3 [Vr–1 – Vr] n n ⇒ Sn = Σ r=1 Tr = 1/3Σ r=1 [Vr–1 – Vr] = 1/3 [V0 – Vn] = 1/3 [1/1.2.3 – 1/(n+1)(n+2)(n+3)] Illustration:
Find the sum of the series 1.2.3.4.5 + 2.3.4.5.6 + …… n terms. Solution:
Let Tr = r(r + 1)(r + 2)(r + 3)(r + 4) = 1/6 [(r+5)–(r–1)] r(r + 1)(r + 2)(r + 3)(r+ 4) = 1/6 [r(r+1)…(r+5)–(r+4)] = [Vr – Vr–1] n ⇒ S = 1/6 Σ r=f [Vr – Vr–1] = 1/6 [Vn – V0] = 1/6[n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)].
Examples based on Relation between AM GM HM
Illustration:
Evaluate: 12 + 22 + 32 + 42 + 52 Solution:
Σ5r=1 r2 = 5(5+1)(2.5+1)/6 = 55 Illustration:
Evaluate: 62 + 72 + 82 + 92 + 102 Solution:
Required sum = 1 2 + 22 + 32 + …… +102 – (12 + 22 +…… 52) = Σ10r=1 r2 – Σ5r=1 r2 = 10(10+1)(2.10+1)/6 – 5(5+1)(2.5+1)6 = 385 – 55 = 330 Illustration:
Evaluate : 12 + 32 + 52 +………+ (2n–1)2 Solution: Note:
There are n in terms in the series.
Required sum = 1 2 + 22 + 32 + 42 +……+ (2n–1)2 + (2n)2 –(22 + 42 +………+ (2n)2) = (2n)(2n+1)(4n+1)/6 – 4 (12 + 22 + ……+ n2) = (2n)(2n+1)(4n+1)/6 – 4 . n(n+1)(2n+1)/6 =n(4n2–1)/3 Illustration:
Sum: 1 . 2 + 2 . 3 + 3 . 4 +…… + n.(n+1) Solution:
tr = r(r+1) = r2 + r t1 = 12 + 1 t2 = 22 + 2 t3 = 32 + 3 tn = n2 + n Adding t1 = t2 + t3 +…+ tn = (12 + 22 + 32 +…+ n2)+(1+2+3+…+n). Σnr=1 tr = Σnr=1 r2 + Σnr=1 r Sn = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)/6 [2n + 1 + 3] = n(n+1)/6 (2n + 4) =n(n+1)(n+2)/3 Illustration:
Sum to n-terms =1/1.2 + 1/2.3 + 1/3.4 +......+ 1/n(n+1) Solution:
tr = 1/r(r+1) = 1/r – 1/r+1 t1 = 1 –1/2 t2 = 1/2 – 1/3 t3 = 1/3 – 1/4 t4 = 1/4 – 1/5 tn =1/n – 1/n+1 Adding
Σnr=1 tr = 1 – 1/n+1 = n/n+1 Factorial:
Factorial of a natural number n is defined as the product of first n natural numbers and it is noted ên or n! |0 = 1
(by definition)
|1 = 1 |2 = 2. |1 = 2 |3 = 3 . |2 = 3.2.1 = 6 |4 = 4 . |3 = 4.3.2.1 =24 |n = n |(n-1) Illustration:
Evaluate the sum : 1 |1 + 2 |2 + 3 |3 +… to n terms. Solution:
tr = r |r = |r + 1 – |r t1 = |2 – |1 t2 = |3 – |2 t3 = |4 – |3 tn = |n + 1 – |n
————————————
Σnr=1 tr = |n + 1 – 1 Illustration:
If x, y, z are positive real numbers, such that x + y + z = a, then prove that 1/x + 1/y + 1/z > 9/a. Solution:
Since A.M. > H.M. x+y+z/3 > 3/x–1+y–1+z–1
⇒
1/x + 1/y + 1/z > 9/a.
Illustration:
Prove that (a + b + c) (ab + bc + ca) > 9abc. Solution:
Using AM > GM, we have a+b+c/3 > (abc)1/3 and ab+bc+ca/3 > (ab.bc.ca)1/3. Multiplying these two results, we have (a+b+c/3)(ab+bc+ca/3) >(abc)1/3 (ab.bc.ca)1/3 3 3 3 1/2
or,(ab+bc+ca)(a+b+c)/9 > (a b c )
or, (a + b + c)(ab + bc + ca) > 9abc.
