Question

In an H.P., ${p^{th}}$ term is $q$ and ${q^{th}}$ term is $p$. Then, the $p{q^{th}}$ term is
A. Zero
B. $\dfrac{{p + q}}{{pq}}$
C. 1
D. $\dfrac{{pq}}{{p + q}}$

Answer 1

Hint: As we know that if $p$ and $q$ are in harmonic progression then \[\dfrac{1}{p}\] and \[\dfrac{1}{q}\] are in arithmetic progression. Then, rewrite the terms in AP by the formula ${a_n} = a + \left( {n - 1} \right)d$. After that, subtract the ${q^{th}}$ term from the ${p^{th}}$ term and simplify to get a common difference. Then, substitute the value of the common difference in any term to get the first term. With these values find the $p{q^{th}}$ term.

Complete step by step answer:
Given that ${p^{th}}$ term is $q$ and ${q^{th}}$ the term is $p$ of an H.P.
A harmonic progression (HP) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic. The progression that does not contain 0. In the HP, any term in the sequence is considered as the Harmonic mean of its two neighbors for example,
The sequence a, b, c, d is considered as an arithmetic progression, the harmonic progression can be written as $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}$.
So, $\dfrac{1}{q}$ is ${p^{th}}$ and $\dfrac{1}{p}$ is ${q^{th}}$ the term of A.P.
Let the A.P. has $a$ as the first term and $d$ as a common difference.
So, the ${p^{th}}$ term will be,
$ \Rightarrow \dfrac{1}{q} = a + \left( {p - 1} \right)d$ ….. (1)
The ${q^{th}}$ term will be,
$ \Rightarrow \dfrac{1}{p} = a + \left( {q - 1} \right)d$ ….. (2)
Subtract equation (2) from equation (1),
$ \Rightarrow \dfrac{1}{q} - \dfrac{1}{p} = a + \left( {p - 1} \right)d - a - \left( {q - 1} \right)d$
Simplify the terms,
$ \Rightarrow \dfrac{{p - q}}{{pq}} = \left( {p - 1 - q + 1} \right)d$
Simplify the terms in the bracket,
$ \Rightarrow \left( {p - q} \right)d = \dfrac{{p - q}}{{pq}}$
Divide both sides by $\left( {p - q} \right)$,
$ \Rightarrow d = \dfrac{1}{{pq}}$
Substitute the value of $d$ in equation (1),
$ \Rightarrow \dfrac{1}{q} = a + \left( {p - 1} \right)\dfrac{1}{{pq}}$
Multiply both sides by $pq$ and simplify it,
$ \Rightarrow p = apq + p - 1$
Simplify the terms,
$ \Rightarrow 1 = apq$
Divide both sides by $pq$,
$ \Rightarrow a = \dfrac{1}{{pq}}$
Now find the $p{q^{th}}$ term of the AP is,
$ \Rightarrow {a_{pq}} = \dfrac{1}{{pq}} + \left( {pq - 1} \right)\dfrac{1}{{pq}}$
Take $\dfrac{1}{{pq}}$ common on the right side,
$ \Rightarrow {a_{pq}} = \dfrac{1}{{pq}}\left( {1 + pq - 1} \right)$
Simplify the terms in the bracket,
$ \Rightarrow {a_{pq}} = \dfrac{1}{{pq}} \times pq$
Cancel out the common terms,
$ \Rightarrow {a_{pq}} = 1$
Thus, the $p{q^{th}}$ term is 1.

Hence, option (C) is the correct answer.

Note: Whenever we come across such problems the key concept is to know the basic definitions of H.P, GP, and AP. It will eventually help you get on the right track to reach the solution as all these have different definitions of three numbers to be in AP, GP, or HP.