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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            =      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}.

 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.

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