Maths > Sequence and Series > 4.0 Harmonic Sequence or Harmonic Progression (H.P.)
Sequence and Series
1.0 Introduction
2.0 Arithmetic Sequence or Arithmetic Progression (A.P.)
3.0 Geometric Sequence or Geometric Progression (G.P.)
4.0 Harmonic Sequence or Harmonic Progression (H.P.)
5.0 Relation among A.M., G.M. and H.M.
6.0 $\Sigma $, Sigma Notation
7.0 Arithmetico-Geometric Series (A.G.S.)
8.0 Miscellaneous Series
4.1 Harmonic mean
Harmonic mean $H$ of any two numbers $a$ and $b$ is $$H = \frac{2}{{\frac{1}{a} + \frac{1}{b}}} = \frac{{2ab}}{{a + b}}$$ where $a$ and $b$ are two non-zero numbers.
If $$a,{H_1},{H_2},...,{H_n},b$$ is in H.P., then $${H_1},{H_2},...,{H_n}$$ are called $n$ harmonic means between $a$ and $b$. If $d$ is the common difference of the corresponding A.P., then $$\begin{equation} \begin{aligned} \frac{1}{b} = \frac{1}{a} + (n + 2 - 1)d \\ \Rightarrow d = \frac{{a - b}}{{ab(n + 1)}} \\\end{aligned} \end{equation} $$
Question 15. If ${A_1},{A_2};{G_1},{G_2};{\text{ and }}{H_1},{H_2}$ be two A.M.s, G.M.s and H.M.s between two quantities $a$ and $b$ then show that $${A_1}{H_2} = {A_2}{H_1} = {G_1}{G_2} = ab$$
Solution: $$a,{A_1},{A_2},b...(1)$$ are in A.P
$$a,{H_1},{H_2},b$$ are in H.P. Therefore, $$\frac{1}{a},\frac{1}{{{H_1}}},\frac{1}{{{H_2}}},\frac{1}{b}...(2)$$ are in A.P.
Multiply equation $(2)$ by $ab$, we get $$b,\frac{{ab}}{{{H_1}}},\frac{{ab}}{{{H_2}}},a...(3)$$ is in A.P.
Comparing $(1)$ and $(3)$, we get $$\begin{equation} \begin{aligned} {A_1} = \frac{{ab}}{{{H_2}}}{\text{ and }}{A_2} = \frac{{ab}}{{{H_1}}} \\ \therefore {A_1}{H_2} = {A_2}{H_1} = ab = {G_1}{G_2} \\\end{aligned} \end{equation} $$
