Solved Examples on Harmonic Progression

Illustration 1: Let ‘a’ and ‘b’ be positive real numbers. If a, A₁, A₂, b are in arithmetic progression, a, G₁, G₂, b are in geometric progression and a, H₁, H₂, b are in harmonic progression, show that

G₁G₂/ H₁H₂ = (A₁ + A₂)/ (H₁ + H₂) = (2a+b)(a+2b)/9ab. (2002)

Solution: We are given that a, A₁, A₂, b are in A.P

Hence, A₁+A₂ = a+b

Now, a, G₁, G₂, b are in G.P

So, G₁G₂ = ab

Similarly, as a, H₁, H₂, b are in H.P

So, H₁ = 3ab/ (2b+a)

And H₂ = 3ab/ (b+2a)

Hence, 1/H₁ + 1/H₂ = 1/a + 1/b

So, (H₁ + H₂)/ H₁.H₂ = (A₁+A₂)/G₁.G₂ = 1/a + 1/b

Now, (G₁G₂)/ H₁.H₂ =

= (2a+b) (a+2b)/ 9ab

Thus combining the obtained results, we get

(G₁G₂)/ H₁.H₂ = (A₁+A₂)/(H₁ + H₂) = (2a+b) (a+2b)/ 9ab

Illustration 2: Let A₁, H₁ denote the arithmetic and harmonic means of two distinct positive numbers. For n ≥ 2, let A_n-1, H_n-1 have arithmetic and harmonic means as A_n and H_n respectively. Then which of the following statements is correct? (2007)

(a) 1. H₁ > H₂ > H₃ > ….

2. H₁ < H₂ < H₃ < ….

3. H₁ > H₃ > H₅ > …. and H₂ < H₄ < H₆ < ….

4. H₁ < H₃ < H₅ < ….  and H₂ > H₄ > H₆ > ….

(b) 1. A₁ > A₂ > A₃ > ….

2. A₁ < A₂ < A₃ < ….

3. A₁ > A₃ > A₅ > …. and A₂ < A₄ < A₆ < ….

4. A₁ < A₃ < A₅ < ….  and A₂ > A₄ > A₆ > ….

Solution: H₁ is the harmonic mean between two numbers ‘a’ and ‘b’.

A₁ = (a+b)/2

H₁ = 2ab/ (a+b)

H_n = 2A_n-1H _n-1 / (A_n-1 + H _n-1)

A_n = (A_n-1 + H _n-1)/ 2

A₂ is A.M of A₁ and H₁ and A₁ > H₁

Hence, A₁ > A₂ > H₁

A₃ is A.M of A₂ and H₂ and A₂ > H₂

So, A₂ > A₃ > H₂

…… …… …….

Hence, A₁ > A₂ > A₃ > ……

Now, as we have stated above A₁ > H₂ > H₁, A₂ > H₃ > H₂

So, H₁ < H₂ < H₃ < ….

Illustration 3: Prove that three quantities a, b, c are in A.P., G.P., or H.P. iff
(a-b)/(b-c) = a/a, a/b or a/c respectively.

Solution: If a, b and c are in A.P, then

b-a = c-b

This can be written as a, b and c are in A.P. iff b-a = c-b

iff (a-b)/(b-c) = 1 = a/a

Similarly, when a, b and c are in G.P,

iff b/a = c/b i.e

iff 1 - b/a = 1 - c/b

i.e. iff (a-b)/a = (b-c)/b

i.e. iff (a-b)/(b-c) = a/b

Similarly a, b, c are in H.P.

iff 1/b - 1/a = 1/c - 1/b

i.e. iff (a-b)/ab = (b-c)/bc

i.e. iff (a-b)/(b-c) = ab/bc = a/c