**Harmonic Progression**

**Insertion of harmonic means between two given numbers: **Let a, b be two given numbers. If n numbers H₁, H₂, … H_{n }are inserted between a and b such that the sequence H₁, H₂, … H _{n ,}b is an HP Then, H₁, H₂, … H_{n} 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, …, A_{n} be n arithmetic means between a and b. Then a, A₁, A₂, …, A_{n}, 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, … H_{n} 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 \({{\left( \frac{n+1}{n-1} \right)}^{2}}p\).

R**elation 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