Question

If the \[{p^{th}},{q^{th}},{r^{th}}\] terms of a H.P. be a, b, c respectively, prove that \[(q - r)bc + (r - p)ca + (p - q)ab = 0\]

Answer 1

Hint: In this question, we have to find the number of terms in the harmonic progression.
First we need to find out the respective terms of a harmonic progression using the $n^{th}$ formula of a H.P., then putting these in the left hand side we will get the required solution.

Formula used: The $n^{th}$ term of the H.P. is represented by \[\dfrac{1}{{x + (n - 1)d}}\]
Where x is the first term, and d is the common difference between the terms and n is the number of terms in the G.P.

Complete step-by-step answer:
It is given that, the \[{p^{th}},{q^{th}},{r^{th}}\] terms of a H.P. be a, b, c respectively.
We need to prove that, \[(q - r)bc + (r - p)ca + (p - q)ab = 0\]
Using the formula for the $n^{th}$ term of the H.P. we get,
\[{a_n} = \dfrac{1}{{x + (n - 1)d}}\], Where x is the first term, and d is the common difference between the terms and n is the number of terms in the G.P.
Thus the \[{p^{th}}\] term of the H.P. is \[{a_p} = \dfrac{1}{{x + (p - 1)d}}\].
We have the \[{p^{th}}\] term of the H.P. is a.
Therefore \[a = \dfrac{1}{{x + (p - 1)d}}\].
That is \[x + \left( {p - 1} \right)d = \dfrac{1}{a}\]………………...i)
The \[{q^{th}}\] term of the H.P. is \[{a_q} = \dfrac{1}{{x + (q - 1)d}}\].
We have the \[{q^{th}}\] term of the H.P. is b.
Therefore \[b = \dfrac{1}{{x + (q - 1)d}}\].
i.e. \[x + \left( {q - 1} \right)d = \dfrac{1}{b}\]..............….ii)
The \[{r^{th}}\] term of the H.P. is \[{a_r} = \dfrac{1}{{x + (r - 1)d}}\].
We have the \[{r^{th}}\] term of the H.P. is c.
Therefore \[c = \dfrac{1}{{x + (r - 1)d}}\].
That is \[x + \left( {r - 1} \right)d = \dfrac{1}{c}\].............…..iii)
Subtracting i) and ii) we get,
$\Rightarrow$\[\left( {p - 1} \right)d - \left( {q - 1} \right)d = \dfrac{1}{a} - \dfrac{1}{b}\]
$\Rightarrow$\[pd - d - qd + d = \dfrac{{b - a}}{{ab}}\]
By cross multiplication we have,
$\Rightarrow$\[ab\left( {p - q} \right)d = b - a\]……………....iv)
Subtracting ii) and iii) we get,
$\Rightarrow$\[\left( {q - 1} \right)d - \left( {r - 1} \right)d = \dfrac{1}{b} - \dfrac{1}{c}\]
$\Rightarrow$\[qd - d - rd + d = \dfrac{{c - b}}{{bc}}\]
By cross multiplication we have,
$\Rightarrow$\[bc\left( {q - r} \right)d = c - b\]……………....v)
Applying iii) and i) we get,
$\Rightarrow$\[\left( {r - 1} \right)d - \left( {p - 1} \right)d = \dfrac{1}{c} - \dfrac{1}{a}\]
$\Rightarrow$\[rd - d - pd + d = \dfrac{{a - c}}{{ac}}\]
By cross multiplication we have,
$\Rightarrow$\[ac\left( {r - p} \right)d = a - c\]…………….....vi)
Adding iv), v), and vi) we get,
$\Rightarrow$\[ab\left( {p - q} \right)d + bc\left( {q - r} \right)d + ac\left( {r - p} \right)d = b - a + c - b + a - c\]
Solving we get, \[\left\{ {(q - r)bc + (r - p)ca + (p - q)ab} \right\}d = 0\].
Since \[d \ne 0\] thus we get, \[(q - r)bc + (r - p)ca + (p - q)ab = 0\].

\[\therefore (q - r)bc + (r - p)ca + (p - q)ab = 0\](proved).

Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-one number called the common ratio. A harmonic progression is a type of progression formed by taking the reciprocals of an arithmetic progression.