9.6 Harmonic Sequence or Harmonic Progression (HP)

A Harmonic Progression (HP) is a sequence, if the reciprocals of its terms are in Arithmetic Progression (AP) i.e., \(t_1, t_2, t_3, \ldots\) is \(HP\) if and only if \(\frac{1}{t_1}, \frac{1}{t_2}, \frac{1}{t_3}, \ldots\) is an AP.

For example, The sequence

(i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \ldots\)
(ii) \(2, \frac{5}{2}, \frac{10}{3}, \ldots\)
(iii) \(\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}, \ldots\) are HP’s.

Remark

• No term of HP can be zero.
• The most general or standard HP is
  \(
  \frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}, \frac{1}{a+3 d}, \cdots
  \)

Example 1: If \(a, b, c\) are in HP, then show that \(\frac{a-b}{b-c}=\frac{a}{c}\).

Solution: Since, \(a, b, c\) are in HP, therefore
\(
\begin{aligned}
& \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text { are in AP } \\
& \text { i.e. } \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b} \\
& \text { or } \frac{a-b}{a b}=\frac{b-c}{b c} \text { or } \frac{a-b}{b-c}=\frac{a}{c}
\end{aligned}
\)

Remarks

• A HP may also be defined as a series in which every three consecutive terms (say \(I, II, III\)) satisfy \(\frac{I-I I}{I I-I I I}=\frac{I}{I I I}\) this relation.

Example 2: Find the first term of a HP whose second term is \(\frac{5}{4}\) and the third term is \(\frac{1}{2}\).

Solution: Let \(a\) be the first term. Then, \(a, \frac{5}{4}, \frac{1}{2}\) are in HP.
Then, \(\frac{a-\frac{5}{4}}{\frac{5}{4}-\frac{1}{2}}=\frac{a}{\frac{1}{2}}\) [from above note]
\(
\begin{aligned}
\frac{4 a-5}{5-2} & =2 a \\
4 a-5=6 a \text { or } 2 a & =-5 \\
a & =-\frac{5}{2}
\end{aligned}
\)

\(n\)th Term of HP from Beginning

Let \(a\) be the first term, \(d\) be the common difference of an \(AP\). Then, \(n\)th term of an AP from beginning \(=a+(n-1) d\) Hence, the \(n\)th term of HP from beginning
\(
=\frac{1}{a+(n-1) d}, \forall n \in N
\)

\(n\)th Term of HP from End

Let \(l\) be the last term, \(d\) be the common difference of an \(AP\). Then, \(n\)th term of an AP from end \(=l-(n-1) d\)
Hence, the \(n\)th term of HP from end \(=\frac{1}{l-(n-1) d}, \forall n \in N\)

Remark

• \(\frac{1}{n \text { thterm of HP from beginning }}+\frac{1}{n \text {th term of HP from end }}\) \(=a+I=\frac{1}{\text { first term of } HP }+\frac{1}{\text { last term of HP }}\)
• There is no general formula for the sum of any number of quantities in \(HP\) are generally solved by inverting the terms and making use of the corresponding AP.

Example 3: If \(\frac{1}{a}+\frac{1}{c}+\frac{1}{a-b}+\frac{1}{c-b}=0\), then prove that \(a, b, c\) are in HP, unless \(b=a+c\).

Solution: We have, \(\frac{1}{a}+\frac{1}{c}+\frac{1}{a-b}+\frac{1}{c-b}=0\)
\(
\begin{aligned}
\Rightarrow & \left(\frac{1}{a}+\frac{1}{c-b}\right)+\left(\frac{1}{c}+\frac{1}{a-b}\right) & =0 \\
\Rightarrow & \frac{(c-b+a)}{a(c-b)}+\frac{(a-b+c)}{c(a-b)} & =0 \\
\Rightarrow & (a+c-b)\left[\frac{1}{a(c-b)}+\frac{1}{c(a-b)}\right] & =0 \\
\Rightarrow & (a+c-b)[2 a c-b(a+c)] & =0
\end{aligned}
\)
If \(\quad a+c-b \neq 0\), then \(2 a c-b(a+c)=0\)
or \(b=\frac{2 a c}{a+c}\)
Therefore, \(a, b, c\) are in HP and if \(2 a c-b(a+c) \neq 0\), then \(a+c-b=0\) i.e., \(b=a+c\).

Example 4: If \(a_1, a_2, a_3, \ldots, a_n\) are in \(HP\), then prove that \(a_1 a_2+a_2 a_3+a_3 a_4+\ldots+a_{n-1} a_n=(n-1) a_1 a_n\)

Solution: Given, \(a_1, a_2, a_3, \ldots, a_n\) are in HP.
\(
\therefore \quad \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots, \frac{1}{a_n} \text { are in AP. }
\)
Let \(D\) be the common difference of the \(AP\), then
\(
\begin{array}{rr}
& \frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{a_3}-\frac{1}{a_2}=\frac{1}{a_4}-\frac{1}{a_3}=\ldots=\frac{1}{a_n}-\frac{1}{a_{n-1}}=D \\
\Rightarrow \quad & \frac{a_1-a_2}{a_1 a_2}=\frac{a_2-a_3}{a_2 a_3}=\frac{a_3-a_4}{a_3 a_4}=\ldots=\frac{a_{n-1}-a_n}{a_{n-1} a_n}=D \\
\Rightarrow \quad & a_1 a_2=\frac{a_1-a_2}{D}, a_2 a_3=\frac{a_2-a_3}{D}, a_3 a_4=\frac{a_3-a_4}{D}, \\
\ldots, a_{n-1} a_n=\frac{a_{n-1}-a_n}{D}
\end{array}
\)
On adding all such expressions, we get
\(
\begin{gathered}
a_1 a_2+a_2 a_3+a_3 a_4+\ldots+a_{n-1} a_n=\frac{a_1-a_n}{D}=\frac{a_1 a_n}{D}\left(\frac{1}{a_n}-\frac{1}{a_1}\right) \\
=\frac{a_1 a_n}{D}\left[\frac{1}{a_1}+(n-1) D-\frac{1}{a_1}\right]=(n-1) a_1 a_n
\end{gathered}
\)
Hence, \(a_1 a_2+a_2 a_3+a_3 a_4+\ldots+a_{n-1} a_n=(n-1) a_1 a_n\)

Remark

In particular case,

• when \(n=4 a_1 a_2+a_2 a_3+a_3 a_4=3 a_4\)
• when \(n=6 a_4 a_2+a_2 a_3+a_3 a_4+a_4 a_5+a_5 a_5=5 a_4 a_5\)

Example 5: The sum of three numbers in HP is 37 and the sum of their reciprocals is \(\frac{1}{4}\). Find the numbers.

Solution: Three numbers in HP can be taken as \(\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}\).
Then, \(\quad \frac{1}{a-d}+\frac{1}{a}+\frac{1}{a+d}=37 \dots(i)\)
\(
\begin{aligned}
\text { and } a-d+a+a+d & =\frac{1}{4} \\
a & =\frac{1}{12}
\end{aligned}
\)
From Eq. (i), \(\frac{12}{1-12 d}+12+\frac{12}{1+12 d}=37\)
\(
\begin{aligned}
& \Rightarrow \frac{12}{1-12 d}+\frac{12}{1+12 d}=25 \Rightarrow \frac{24}{1-144 d^2}=25 \\
& \Rightarrow \quad 1-144 d^2=\frac{24}{25} \text { or } d^2=\frac{1}{25 \times 144} \\
& \therefore \quad d= \pm \frac{1}{60} \\
& \therefore a-d, a, a+d \text { are } \frac{1}{15}, \frac{1}{12}, \frac{1}{10} \text { or } \frac{1}{10}, \frac{1}{12}, \frac{1}{15} .
\end{aligned}
\)
Hence, three numbers in HP are \(15,12,10\) or \(10,12,15\).

Example 6: If \(p\) th, \(q\) th and \(r\) th terms of a HP be respectively \(a, b\) and \(c\), then prove that
\(
(q-r) b c+(r-p) c a+(p-q) a b=0 .
\)

Solution: Let \(A\) and \(D\) be the first term and common difference of the corresponding AP. Now, \(a, b, c\) are respectively the \(p\) th, \(q\) th and \(r\) th terms of HP. \(\therefore \frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) will be respectively the \(p\) th, \(q\) th and \(r\) th terms of the corresponding \(AP\).
\(
\begin{aligned}
& \frac{1}{a}=A+(p-1) D \dots(i) \\
& \frac{1}{b}=A+(q-1) D \dots(ii) \\
& \frac{1}{c}=A+(r-1) D \dots(iii)
\end{aligned}
\)
On subtracting Eq. (iii) from Eq. (ii), we get
\(
\frac{1}{b}-\frac{1}{c}=(q-r) D \Rightarrow b c(q-r)=\frac{(c-b)}{D}=-\frac{(b-c)}{D}
\)
So, LHS \(=(q-r) b c+(r-p) c a+(p-q) a b\)
\(
=-\frac{1}{D}\{b-c+c-a+a-b\}=0=\text { RHS }
\)

Theorem Relating to the Three Series

• If \(a, b, c\) are three consecutive terms of a series, then if \(\frac{a-b}{b-c}=\frac{a}{a}\), then \(a, b, c\) are in AP.
• If \(\frac{a-b}{b-c}=\frac{a}{b}\), then \(a, b, c\) are in GP and
• If \(\frac{a-b}{b-c}=\frac{a}{c}\), then \(a, b, c\) are in HP.

Example 7: If \(a, b, c\) are in AP and \(a^2, b^2, c^2\) be in \(H P\). Then, prove that \(-\frac{a}{2}, b, c\) are in GP or else \(a=b=c\).

Solution: Given, \(a, b, c\) are in AP.
\(
\therefore \quad b=\frac{a+c}{2} \dots(i)
\)
and \(a^2, b^2, c^2\) are in HP.
\(
\therefore \quad b^2=\frac{2 a^2 c^2}{a^2+c^2} \dots(ii)
\)
From Eq. (ii) \(b^2\left\{(a+c)^2-2 a c\right\}=2 a^2 c^2\)
\(
\Rightarrow \quad b^2\left\{(2 b)^2-2 a c\right\}=2 a^2 c^2 \text { [from Eq. (i)] }
\)
\(
\begin{array}{lr}
\Rightarrow & 2 b^4-a c b^2-a^2 c^2=0 \\
\Rightarrow & \left(2 b^2+a c\right)\left(b^2-a c\right)=0 \\
\Rightarrow & 2 b^2+a c=0 \text { or } b^2-a c=0
\end{array}
\)
If \(2 b^2+a c=0\), then \(b^2=-\frac{1}{2} a c\) or \(-\frac{a}{2}, b, c\) are in GP and if \(b^2-a c=0 \Rightarrow a, b, c\) are in GP.
But given, \(a, b, c\) are in AP.
Which is possible only when \(a=b=c\)

Example 8: If \(a, b, c\) are in HP, b,c,d are in GP and \(c, d, e\) are in AP, then show that \(e=\frac{a b^2}{(2 a-b)^2}\).

Solution: Given, \(a, b, c\) are in HP.
\(
\therefore \quad b=\frac{2 a c}{a+c} \text { or } c=\frac{a b}{2 a-b} \dots(i)
\)
Given, \(b, c, d\) are in GP.
\(
\therefore \quad c^2=b d \dots(ii)
\)
and given, \(c, d, e\) are in AP.
\(
\begin{array}{rlrl}
\therefore & d & =\frac{c+e}{2} \\
\Rightarrow & e & =2 d-c \\
e & =\left(\frac{2 c^2}{b}-c\right) \text { [from Eq. (ii)] …(iii) }
\end{array}
\)
From Eqs. (i) and (iii), \(e=\frac{2}{b}\left(\frac{a b}{2 a-b}\right)^2-\left(\frac{a b}{2 a-b}\right)\)
\(
\begin{aligned}
& =\frac{a b}{(2 a-b)^2}\{2 a-(2 a-b)\} \\
& =\frac{a b^2}{(2 a-b)^2}
\end{aligned}
\)

Example 9: If \(a, b, c, d\) and \(e\) be five real numbers such that \(a, b, c\) are in AP; \(b, c, d\) are in GP; c, \(d, e\) are in HP. If \(a=2\) and \(e=18\), then find all possible values of \(b, c\) and \(d\).

Solution: Given, \(a, b, c\) are in AP,
\(
\therefore \quad b=\frac{a+c}{2} \dots(i)
\)
\(b, c, d\) are in GP,
\(
\therefore \quad c^2=b d \dots(ii)
\)
and \(c, d, e\) are in \(HP\).
\(
\therefore \quad d=\frac{2 c e}{c+e} \dots(iii)
\)
Now, substituting the values of \(b\) and \(d\) in Eq. (ii), then
\(
\begin{aligned}
& c^2=\left(\frac{a+c}{2}\right)\left(\frac{2 c e}{c+e}\right) \\
& \Rightarrow \quad c(c+e)=e(a+c) \\
& \Rightarrow \quad c^2=a e \dots(iv) \\
&
\end{aligned}
\)
Given, \(\quad a=2, e=18\)
From Eq. (iv), \(\quad c^2=(2)(18)=36\)
\(\therefore \quad c= \pm 6\)
From Eq. (i), \(\quad b=\frac{2 \pm 6}{2}=4,-2\)
and from Eq. (ii), \(\quad d=\frac{c^2}{b}=\frac{36}{b}=\frac{36}{4}\) or \(\frac{36}{-2}\)
\(\therefore \quad d=9\) or -18
Hence, \(\quad c=6, b=4, d=9\) or \(c=-6, b=-2, d=-18\)

Example 10: If three positive numbers \(a, b\) and \(c\) are in AP, GP and HP as well, then find their values.

Solution: Since \(a, b, c\) are in AP, GP and HP as well
\(
b=\frac{a+c}{2} \dots(i)
\)
\(
b^2=a c \dots(ii)
\)
and \(b=\frac{2 a c}{a+c} \dots(iii)\)
From Eqs. (i) and (ii), we have
\(
\left(\frac{a+c}{2}\right)^2=a c
\)
\(
\begin{aligned}
(a+c)^2 & =4 a c \\
(a+c)^2-4 a c & =0 \\
(a-c)^2 & =0 \\
a & =c \dots(iv)
\end{aligned}
\)
On putting \(c=a\) in Eq. (i), we get \(b=\frac{a+a}{2}=a \dots(v)\)
From Eqs. (iv) and (v), \(a=b=c\), thus the three numbers will be equal.

Remark

• If three positive numbers are in any two of \(AP , GP\) and \(HP\), then it will be also in third.
• Thus, if three positive numbers are in any two of AP, GP and \(HP\), then they will be in the third progression and the numbers will be equal.