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