Question

The ${p^{th}}$ , ${q^{th}}$ and ${r^{th}}$ term of a HP are $a,b$ and $c$ respectively. Then prove that $\dfrac{{q - r}}{a} + \dfrac{{r - p}}{b} + \dfrac{{p - q}}{c} = 0$

Answer 1

If x,y and z are in harmonic progression then $\dfrac{1}{x},\dfrac{1}{y}$ and $\dfrac{1}{z}$ are in arithmetic progression.

Complete step-by-step Solution:
In order to proceed with the proof, the value of individual terms $\left( {\dfrac{{q - r}}{a},\dfrac{{r - p}}{b},\dfrac{{p - q}}{c}} \right)$ in the expression is to be calculated and their sum is to be evaluated.
The ${p^{th}}$ , ${q^{th}}$ and ${r^{th}}$ term of a HP are $a,b$ and $c$ respectively.
It means that $\dfrac{1}{a},\dfrac{1}{b}$ and $\dfrac{1}{c}$ are in arithmetic progression.
For an AP the , ${n^{th}}$ term is given by,
${a_n} = a + \left( {n - 1} \right)d$
Where,
$a =$ First term
$d =$ Common difference
$n =$ Number of terms in an AP
Let’s suppose the first term of an arithmetic progression is ${a_1}$ and the common difference, ${d_1}$ .${p^{th}}$ term of an arithmetic progression is given by,
${a_p} = {a_1} + \left( {p - 1} \right){d_1} \cdots \left( 1 \right)$
But the ${p^{th}}$ term is $\dfrac{1}{a}$ , substitute it in equation (1)
$\Rightarrow \dfrac{1}{a} = {a_1} + \left( {p - 1} \right){d_1} \cdots \left( 2 \right)$
${q^{th}}$ term of an arithmetic progression is given by,
$\Rightarrow{a_q} = {a_1} + \left( {q - 1} \right){d_1} \cdots \left( 3 \right)$
But the ${q^{th}}$ term is $\dfrac{1}{b}$ , substitute it in equation (3) we get
$\Rightarrow\dfrac{1}{b} = {a_1} + \left( {q - 1} \right){d_1} \cdots \left( 4 \right)$
${r^{th}}$ term of an arithmetic progression is given by,
$\Rightarrow{a_r} = {a_1} + \left( {r - 1} \right){d_1} \cdots \left( 5 \right)$
But the ${r^{th}}$ term is $\dfrac{1}{c}$ , substitute it in equation (5)
$\Rightarrow\dfrac{1}{c} = {a_1} + \left( {r - 1} \right){d_1} \cdots \left( 6 \right)$

Now, the value of the terms $\left( {\dfrac{{q - r}}{a} + \dfrac{{r - p}}{b} + \dfrac{{p - q}}{c}} \right)$ should be calculated and their sum should be evaluated.
The term $\left( {\dfrac{{q - r}}{a}} \right)$ can be written as
$\Rightarrow\dfrac{{q - r}}{a} = \left( {q - r} \right) \cdot \dfrac{1}{a} \cdots \left( 7 \right)$
Substitute the value of $\dfrac{1}{a}$ from equation (2) in equation (7)
$\Rightarrow\dfrac{{q - r}}{a} = \left( {q - r} \right) \cdot \left( {{a_1} + \left( {p - 1} \right){d_1}} \right) \cdots \left( 8 \right)$
The term $\left( {\dfrac{{r - p}}{b}} \right)$ can be written as
$\Rightarrow\dfrac{{r - p}}{b} = \left( {r - p} \right) \cdot \dfrac{1}{b} \cdots \left( 9 \right)$
Substitute the value of $\dfrac{1}{b}$ from equation (4) in equation (9)
$\Rightarrow\dfrac{{r - p}}{b} = \left( {r - p} \right) \cdot \left( {{a_1} + \left( {q - 1} \right){d_1}} \right) \cdots \left( {10} \right)$
The term $\left( {\dfrac{{p - q}}{c}} \right)$ can be written as
$\Rightarrow\dfrac{{p - q}}{c} = \left( {p - q} \right) \cdot \dfrac{1}{c} \cdots \left( {11} \right)$
Substitute the value of $\dfrac{1}{c}$ from equation (6) in equation (11)
$\Rightarrow\dfrac{{p - q}}{c} = \left( {p - q} \right) \cdot \left( {{a_1} + \left( {r - 1} \right){d_1}} \right) \cdots \left( {12} \right)$
On Adding equation (8), (10) and (12), we get it as
$\Rightarrow\dfrac{{q - r}}{a} + \dfrac{{r - p}}{b} + \dfrac{{p - q}}{c} = \left( {q - r} \right) \cdot \left( {{a_1} + \left( {p - 1} \right){d_1}} \right) + \left( {r - p} \right) \cdot \left( {{a_1} + \left( {q - 1} \right){d_1}} \right) + \left( {p - q} \right) \cdot \left( {{a_1} + \left( {r - 1} \right){d_1}} \right) \cdots \left( {10} \right)$
Now solving the above expression by opening the terms in the bracket,

$$RHS = \left( {q - r} \right) \cdot \left( {{a_1} + \left( {p - 1} \right){d_1}} \right) + \left( {r - p} \right) \cdot \left( {{a_1} + \left( {q - 1} \right){d_1}} \right) + \left( {p - q} \right) \cdot \left( {{a_1} + \left( {r - 1} \right){d_1}} \right) \\\ = q{a_1} + qp{d_1} - q{d_1} - r{a_1} - rp{d_1} + r{d_1} + r{a_1} + rq{d_1} - r{d_1} - p{a_1} - pq{d_1} + p{d_1} + p{a_1} + pr{d_1} - p{d_1} - q{a_1} - qr{d_1} + q{d_1} \\\ = 0 \\\$$

It is interesting to see that every term in the expression consists of its additive inverse.
Therefore, every term cancels out.
Therefore, $\dfrac{{q - r}}{a} + \dfrac{{r - p}}{b} + \dfrac{{p - q}}{c} = 0$.
Hence, it is proved.

Note: The important step is the starting step which tells that if and are in Harmonic progression then $\dfrac{1}{a},\dfrac{1}{b}$ and$\dfrac{1}{c}$ are in arithmetic progression
Some important properties of harmonic progression are

If a and b are two non-zero numbers, then the harmonic mean of a and b is a number say H such that the sequence consisting of a, H and b are in harmonic progression.
$G = \dfrac{{2ab}}{{a + b}}$
If ${a_1},{a_2} \ldots {a_n}$ are n non-zero numbers, then the harmonic mean G of these numbers is given by,
$\dfrac{1}{H} = \dfrac{{\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + \ldots + \dfrac{1}{{{a_2}}}}}{n}$ .