Harmonic Progression (HP) - Definition, Formulas and Examples

A Harmonic Progression (HP) is defined as a sequence of real numbers obtained by taking the reciprocals of an Arithmetic Progression that excludes 0. In simpler terms, a sequence $a_1, a_2, a_3, \ldots ., a_n, \ldots$ of non-zero numbers is called a harmonic progression if the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots ., \frac{1}{a_n}, \ldots$ is an arithmetic progression. In real life, we use Harmonic Progression (HP) in electrical gadgets, generation of power, used in the field of music to prepare notes.

In this article, we will cover the concept of Harmonic Progression. This category falls under the broader category of Matrices, which is a crucial Chapter in class 11 Mathematics. It is essential not only for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main Exam (from 2013 to 2023), a total of nine questions have been asked on this concept.

Harmonic Progression

A Harmonic Progression (HP) is defined as a sequence of real numbers obtained by taking the reciprocals of an Arithmetic Progression that excludes 0.

A sequence $a_1, a_2, a_3, \ldots ., a_n, \ldots$ of non-zero numbers is called a harmonic progression if the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots, \frac{1}{a_n}, \ldots$.

OR
Reciprocals of arithmetic progression is a Harmonic progression.
E.g. $\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots .$. is an HP because their reciprocals $2,5,8,11, \ldots$ form an A.P.

• No term of the H.P. can be zero.

• The general form of HP is

$\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}, \frac{1}{a+3 d} \ldots$

Here a is the first term and d is the common difference of corresponding A.P.

The general term of a Harmonic Progression

The nth term or general term of an H.P. is the reciprocal of the nth term of the corresponding A.P.
Thus, if $a_1, a_2, a_3, \ldots \ldots, a_n$ is an H.P. and the common difference of corresponding A.P. is d, i.e. $d=\frac{1}{a_n}-\frac{1}{a_{n-1}}$, then the nth term of corresponding AP is $\frac{1}{a_1}+(n-1) d$, and hence, the
general term or nth term of an H.P. is given by

$
a_n=\frac{1}{\frac{1}{a_1}+(n-1) d}
$

Note:

There is no general formula for the sum of n terms that are in H.P.

Recommended Video Based on Harmonic Progression:

Solved Examples Based on Harmonic Progression

Example 1: If $a_1, a_2 \ldots ., a_n$ are in H.P., then the expression $a_1 a_2+a_2 a_3+\ldots . .+a_{n-1} a_n$ is equal to:

Solution: Let the common difference of the corresponding AP be d

So, $\frac{1}{a_2}-\frac{1}{a_1}=d$

$
a_1-a_2=d a_1 a_2
$
Similarly,

$
a_2-a_3=d a_2 a_3
$
$
\begin{aligned}
& a_{n-2}-a_{n-1}=d a_{n-1} a_{n-2} \\
& a_{n-1}-a_n=d a_{n-1} a_n
\end{aligned}
$
Adding all the equations,

$
a_1-a_n=d\left(a_1 a_2+a_2 a_3+\ldots \ldots \ldots \ldots a_{n-1} a_n\right)
$
Now, $\frac{1}{a_n}=\frac{1}{a_1}+(n-1) d$

$
a_1-a_n=(n-1) a_1 a_n d
$
So from (i) \& (ii) on comparison

$
\begin{aligned}
& d\left(a_1 a_2+a_2 a_3+\ldots \ldots \ldots \ldots a_{n-1} a_n\right)=(n-1) a_1 a_n d \\
& a_1 a_2+a_2 a_3+\ldots \ldots \ldots \ldots a_{n-1} a_n=(n-1) a_1 a_n
\end{aligned}
$
So, the answer is $(n-1) a_1 a_n$.
Hence, the answer is $(n-1) a_1 a_n$

Example 2: If $x, y, z$ are in HP, then $\log (x+z)+\log (x-2 y+z)$ is equal to
Solution: We know, The general term of HP is
$
\begin{aligned}
& T_n=\frac{1}{a+(n-1) d} \\
& \text { where } a=\frac{1}{a_1} \text { and } d=\frac{1}{a_2}-\frac{1}{a_1}
\end{aligned}
$
$
\begin{aligned}
& \mathrm{y}=\frac{2 x z}{x+z} \\
& \log (\mathrm{x}+\mathrm{z})+\log (\mathrm{x}-2 \mathrm{y}+\mathrm{z}) \\
& =\log (\mathrm{x}+\mathrm{z}) \cdot(\mathrm{x}-2 \mathrm{y}+\mathrm{z}) \\
& =\log \left[(\mathrm{x}+\mathrm{z})^2-2 \mathrm{y}(\mathrm{x}+\mathrm{z})\right] \\
& =\log \left((x+z)^2-2 y \frac{2 x z}{y}\right) \\
& =\log \left[(\mathrm{x}+\mathrm{z})^2-4 \mathrm{xz}\right]=\log (\mathrm{x}-\mathrm{z})^2=2 \log (\mathrm{x}-\mathrm{z})
\end{aligned}
$
Hence, the answer is $2 \log (x-z)$

Example 3 : If $a_1, a_2, a_3, \ldots \ldots, a_n$ are in H.P then $a_1 a_2+a_2 a_3+\ldots \ldots \ldots+a_{n-1} a_n$ will be equal to :

Solution: the general term or nth term of an H.P. is given by

$
a_n=\frac{1}{\frac{1}{a_1}+(n-1) d}
$

Now,

$
\begin{aligned}
& a_1, a_2, a_3, \ldots \ldots, a_n \text { are in H.P. } \\
& \frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{a_3}-\frac{1}{a_2} \ldots \ldots .=\frac{1}{a_n}-\frac{1}{a_{n-1}}=d \\
& \frac{a_1-a_2}{a_1 \cdot a_2}=\frac{a_2-a_3}{a_2 \cdot a_3} \ldots .=\frac{a_{n-1}-a_n}{a_n \cdot a_{n-1}}=d \\
& a_1-a_2=d a_1 \cdot a_2, \quad a_2-a_3=d a_2 \cdot a_3, \quad \ldots \cdot a_{n-1}-a_n=d a_n \cdot a_{n-1}
\end{aligned}
$
Adding all these

$
\begin{aligned}
& a_1-a_2+a_2-a_3+\ldots+a_{n-1}-a_n=d\left(a_1 \cdot a_2+a_2 \cdot a_3+\ldots+a_{n-1}\right. \\
& a_1-a_n=d\left(a_1 \cdot a_2+a_2 \cdot a_3+\ldots+a_{n-1} \cdot a_n\right) \ldots(i)
\end{aligned}
$

nth term of H.P.

$
\begin{aligned}
& \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d \\
& \frac{1}{a_n}-\frac{1}{a_1}=(n-1) d \\
& \frac{a_1-a_n}{a_1 a_n}=(n-1) d
\end{aligned}
$

by equation (i) and (ii)

$
\left(a_1 \cdot a_2+a_2 \cdot a_3+\ldots+a_{n-1} \cdot a_n\right)=(n-1) a_1 a_n
$
Hence, the answer is $(n-1) a_1 a_n$

Example 4: If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P. and $a-b, c-a, b-c$ are in H.P., then the value of $a+4 b+c$ is equal to

Solution: Given $a, b, c$ are in G.P
Let $r$ be the common ratio

$
b=a r, c=a r^2
$
Now $(a-b),(c-a),(b-c)$ are in H.P.

$\frac{1}{a-b}, \frac{1}{c-a}, \frac{1}{b-c}$ are in A.P.
$\Rightarrow(b-c),(c-a),(a-b),(b-c),(a-b),(c-a)$ are in A.P.
Now $\left(a r-a r^2\right)\left(a r^2-a\right),(a-a r)\left(a r-a r^2\right),(a-a r)\left(a r^2-a\right)$ are in A.P.

$\begin{aligned}
& \left(r-r^2\right)\left(r^2-1\right),(1-r)\left(r-r^2\right),(1-r)\left(r^2-1\right) \text { are in A.P. } \\
& r(1-r)(r-1)(r+r),(1-r) r(1-r),(1-r)(r-1),(r+1) \text { are in A.P. } \\
& r(1+r),-r(r+1) \text { are in A.P } \\
& r(1-r)(r-1)(r+r),(1-r) r(1-r)(r-1),(r+1) \text { are in A.P. } \\
& r(1+r),-r,(r+1) \text { are in A.P. }
\end{aligned}$

Now,

$\begin{aligned} & -2 r=r(1+r)+(1+r) \\ & \Rightarrow-2 r=(1+r)(1+r)\end{aligned}$

$\begin{aligned}
&\Rightarrow r^2+4 r+1=0\\
&\ldots \text { (i) }\\
&\text { Now value of } a+4 b+c
\end{aligned}$

$\begin{aligned} a+4(a r)+\left(a r^2\right)= & a\left(1+4 r+r^2\right) \\ & =a \times 0=0\end{aligned}$

Hence, the answer is 0

Example 5: If $\mathrm{a}_{\mathrm{p}}, \mathrm{a}_{\mathrm{q}}, \mathrm{a}_{\mathrm{x}}$ of an HP are $\mathrm{a}, \mathrm{b}$, and c respectively then the value of $(q-r) b c+(r-p) a c+(p-q) a b_{\text {is }}$

Solution: Let the first term be a and the common difference be $\$ \mathrm{~d} \$ for the corresponding AP.

So,

$
a+(p-1) d=\frac{1}{a_p}
$

$\begin{aligned}
&\Rightarrow a+(p-1) d=\frac{1}{a_p}\\
&\text { ......(1) }\\
&\text { Similarly, }\\
&a+(q-1) d=\frac{1}{b}\\
&\text { (2) }
\end{aligned}$

$\begin{aligned}
&\text { and }\\
&a+(r-1) d=\frac{1}{c}\\
&\ldots . .(3)
\end{aligned}$

$(1)-(2)$

$
\begin{aligned}
& (p-q) d=\frac{1}{a}-\frac{1}{b} \\
& \Rightarrow(p-q) d=\frac{b-a}{a b} \\
& \Rightarrow(p-q) a b=\frac{b-a}{d}
\end{aligned}
$
Similarly,

$
(q-r) b c=\frac{(c-d)}{d}
$

and

$\begin{aligned}
&(r-p) a c=\frac{(a-c)}{d}\\
&\text { .....(6) }
\end{aligned}$

$\begin{aligned}
& (4)+(5)+(6) \\
& (p-q) a b+(q-r) b c+(r-p) a c=\frac{b-a+c-b+a-c}{d}=0
\end{aligned} $

Hence, the answer is 0