Important Mathematics Notes: Harmonic Progression

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

Example 1:

Find the 4^th and 8^th term of the series 6, 4, 3, ……

Solution:

Consider 1/6, /14, 1/3, ...... ∞

Here T₂ – T₁ = T₃ – T₂ = 1/12 ⇒ 1/6, 1/4, 1/3 is an A.P.

4^th term of this A.P. = 1/6 + 3 × 1/12 = 1/6 + 1/4 = 5/12,

And the 8^th term = 1/6 + 7 × 1/12 = 9/12.

Hence the 8^th term of the H.P. = 12/9 = 4/3 and the 4^th term = 12/5.

Example 2:

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.

Example 1:

Find the nth term and the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 +…

Solution:

r^th term of the series = r(r+1).(r+3)=r³ + 4r² + 3r

So sum of n terms = Σⁿ_r=1 r³ + 4Σⁿ_r=1 r² + 3Σⁿ_r=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 2:

Find the sum of the series 1.n + 2(n–1) + 3.(n–2) +…+ n.1.

Solution:

The r^th term of the series is

t_r = (1 + (r – 1).1)(n + (r–1)(–1))

= r(n – r + 1) = r(n + 1) – r²

⇒ S_n= Σⁿ_r=1 t_r Σⁿ_r=1 (n+1)r – Σⁿ_r=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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