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

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$$$\left( {p - 1} \right)d - \left( {q - 1} \right)d = \dfrac{1}{a} - \dfrac{1}{b}$$ \Rightarrowpd - d - qd + d = \dfrac{{b - a}}{{ab}}$$ By cross multiplication we have, $\Rightarrowab\left( {p - q} \right)d = b - a……………....iv) Subtracting ii) and iii) we get, $\Rightarrow$$$\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, \Rightarrowbc\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.