# Harmonic Progression

## Harmonic Progression

Insertion of harmonic means between two given numbers: Let a, b be two given numbers. If n numbers H₁, H₂, … Hn are inserted between a and b such that the sequence H₁, H₂, … H n ,b is an HP Then,  H₁, H₂, … Hn are called n harmonic means between a and b.

Harmonic mean of the given numbers: If a and b are two non-zero numbers. Then the harmonic means of a and b is a number H such that the sequence a, H, b is a H.P.

Now, a, H, b is a HP.

$$\frac{1}{a},\frac{1}{H},\frac{1}{b}$$ is an AP.

$$\frac{2}{H}=\frac{1}{a}+\frac{1}{b}\Rightarrow H=\frac{2ab}{a+b}$$

The harmonic mean H between two numbers a and b is given by $$H=\frac{2ab}{a+b}$$

If a₁, a₂, … a n are n non-zero numbers then the harmonic mean H of these numbers is given by $$\frac{1}{H}=\frac{\frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+……+\frac{1}{{{a}_{n}}}}{n}$$

Example: If p be the first of n arithmetic means between two numbers and q be the first of n harmonic means between the same two numbers, prove that the value of q cannot be between p and $${{\left( \frac{n+1}{n-1} \right)}^{2}}p$$.

Solution: Let the two positive numbers be a and b. Let A₁, A₂, …, An be n arithmetic means between a and b. Then a, A₁, A₂, …, An, b is an AP with common difference $$d=\frac{b-a}{n+1}$$.

P = A₁ = a + d = a + (b – a)/(n+1)P

⇒$$p=\frac{na+b}{n+1}$$ … (1)

Let H₁, H₂, … Hn be n harmonic means between a and b. Then, $$\frac{1}{a},\frac{1}{{{H}_{1}}},\frac{1}{{{H}_{2}}},\,…\,,\frac{1}{{{H}_{n}}},\frac{1}{b}$$ is an AP. With common difference

$$D=\frac{a-b}{\left( n+1 \right)ab}$$

∴ $$\frac{1}{q}=\frac{1}{a}+D$$

⇒ $$\frac{1}{q}=\frac{1}{a}+\frac{a-b}{\left( n+1 \right)ab}\Rightarrow \frac{1}{q}=\frac{nb+a}{\left( n+1 \right)ab}$$

⇒$$q=\frac{\left( n+1 \right)ab}{nb+a}$$ … (2)

From (i), we get

b = (n+1) p – na

Putting the value of b in (ii), we get

q {n (n + 1) p – n²a + a} = (n + 1)a {(n + 1)p – na}

n(n + 1)a² – {(n + 1)²p + (n² – 1)q}a + n(n + 1)pq = 0

na² – {(n + 1) p + (n² – 1)q} + npq = 0

Since a is real, therefore

{(n + 1)p + (n – 1)q}² – 4n²pq  ≥ 0

(n + 1)²p² + (n – 1)²q² + 2(n² – 1)pq – 4n²pq  ≥ 0

(n + 1)²p² + (n – 1) ²q² – 2(n² + 1) pq  ≥ 0

⇒ $${{q}^{2}}-\frac{2\left( {{n}^{2}}+1 \right)}{{{\left( n-1 \right)}^{2}}}pq+{{\left( \frac{n+1}{n-1} \right)}^{2}}{{p}^{2}}\ge 0$$

⇒$${{q}^{2}}-\left\{ 1+{{\left( \frac{n+1}{n-1} \right)}^{2}} \right\}pq+{{\left( \frac{n+1}{n-1} \right)}^{2}}{{p}^{2}}\ge 0$$

⇒$$\left( q-p \right)\left\{ q-{{\left( \frac{n+1}{n-1} \right)}^{2}}p \right\}\ge 0$$

⇒ q < p or $$q>{{\left( \frac{n+1}{n-1} \right)}^{2}}p\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because {{\left( \frac{n+1}{n-1} \right)}^{2}}p>p \right]$$

Hence q cannot lie between p and $${{\left( \frac{n+1}{n-1} \right)}^{2}}p$$.

Relation between Arithmetic, Geometric and Harmonic means between two given numbers:

Let A, G and H be arithmetic, geometric and harmonic means of two positive numbers a and b. Then $$A=\frac{a+b}{2}$$, $$G=\sqrt{ab}$$ and $$H=\frac{2ab}{a+b}$$.

A ≥ G > H

A, G, H form a G.P., i.e., G² = AH