Harmonic Progression (HP) - Definition, Formulas and Examples

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