Harmonic Mean in HP - Practice Questions & MCQ

Harmonic Mean

$
H=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\ldots+\frac{1}{a_n}}
$
If $a_1, a_2, a_3, \ldots ., a_n$ are $n$ positive numbers, then the Harmonic Mean of these numbers is given by If $a$ and $b$ are two numbers and $H$ is the $H M$ of $a$ and $b$. Then, $a, H, b$ are in harmonic progression. Hence,

$
\mathrm{H}=\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2 a b}{a+b}
$
Note that if the AM betw
$\frac{\frac{1}{a}+\frac{1}{b}}{2}$ i.e. $\frac{2 a b}{a+b}$.

Insertion of $\mathbf{n}$-Harmonic Mean Between $\mathbf{a}$ and $\mathbf{b}$
Let $\mathrm{H}_1, \mathrm{H}_2, \mathrm{H}_3 \ldots, \mathrm{H}_{\mathrm{n}}$ be n harmonic mean between two numbers a and b. Then, $a, \mathrm{H}_1, \mathrm{H}_2, \mathrm{H}_3 \ldots, \mathrm{H}_{\mathrm{n}}, b$ is in $\mathrm{H} . \mathrm{P}$.
Hence, $\frac{1}{a}, \frac{1}{\mathrm{H}_1}, \frac{1}{\mathrm{H}_2}, \ldots, \frac{1}{\mathrm{H}_{\mathrm{n}}}, \frac{1}{b}$ are in A.P.

This H.P. contains $\mathrm{n}+2$ terms.
Let, D be the common difference of this A.P. Then,

$
\begin{aligned}
& \therefore \frac{1}{\mathrm{~b}}=(n+2)^{t h} \text { term of } \mathrm{AP} \\
& \Rightarrow \frac{1}{\mathrm{~b}}=\frac{1}{\mathrm{a}}+(\mathrm{n}+1) \mathrm{D} \\
& \Rightarrow \mathrm{D}=\frac{\mathrm{a}-\mathrm{b}}{(\mathrm{n}+1) \mathrm{ab}}
\end{aligned}
$
 

Important Property of HM
The sum of reciprocals of n harmonic means between two numbers is n times the reciprocal of a single H.M. between them.
Proof:
Let $\mathrm{H}_1, \mathrm{H}_2, \mathrm{H}_3 \ldots, \mathrm{H}_{\mathrm{n}}$ be n harmonic means between two numbers a and b. Then, $a, \mathrm{H}_1, \mathrm{H}_2, \mathrm{H}_3 \ldots, \mathrm{H}_{\mathrm{n}}, b$ is an H.P.

$
\begin{aligned}
\therefore \frac{1}{\mathrm{H}_1} & +\frac{1}{\mathrm{H}_2}+\frac{1}{\mathrm{H}_3}+\ldots+\frac{1}{\mathrm{H}_{\mathrm{n}}}=\frac{\mathrm{n}}{2}\left(\frac{1}{\mathrm{H}_1}+\frac{1}{\mathrm{H}_{\mathrm{n}}}\right) \\
& =\frac{\mathrm{n}}{2}\left(\frac{1}{\mathrm{a}}+\mathrm{D}+\frac{1}{\mathrm{~b}}-\mathrm{D}\right)=\frac{\mathrm{n}}{2}\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}\right) \\
& =\frac{\mathrm{n}}{[\text { H.M. of } a \text { and } b]}
\end{aligned}
$