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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.

 

Example : 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, ..................

 

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

Example

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.

 

Example

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.

 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.

Example

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 {3n2 + 19n + 26}.

 

Example

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

 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

              =((√a)+(√b)–2√a√b)/2 > 0

 i.e. AM > GM

 
Similarly, 
G.M. – H.M. = √ab –2ab/a+b > 0

So. GM > HM

 
Combining both results, we get

 AM > GM > HM                                                   …….. (12)

Some other relevant sequences  

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

Example

Evaluate: 12 + 22 + 32 + 42 + 52

Solution

Σ5r=1 r2 = 5(5+1)(2.5+1)/6 = 55

 

Example

Evaluate: 62 + 72 + 82 + 92 + 102

Solution

Required sum = 12 + 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

 

Example

Evaluate : 12 + 32 + 52 +………+ (2n–1)2

Solution

Note:
          There are n in terms in the series.
          Required sum = 12 + 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

 

Example

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

 

Example

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





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