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Question: In an H.P., \({p^{th}}\) term is \(q\) and \({q^{th}}\) term is \(p\). Then, the \(p{q^{th}}\) term ...

In an H.P., pth{p^{th}} term is qq and qth{q^{th}} term is pp. Then, the pqthp{q^{th}} term is
A. Zero
B. p+qpq\dfrac{{p + q}}{{pq}}
C. 1
D. pqp+q\dfrac{{pq}}{{p + q}}

Explanation

Solution

As we know that if pp and qq are in harmonic progression then 1p\dfrac{1}{p} and 1q\dfrac{1}{q} are in arithmetic progression. Then, rewrite the terms in AP by the formula an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d. After that, subtract the qth{q^{th}} term from the pth{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 pqthp{q^{th}} term.

Complete step by step answer:
Given that pth{p^{th}} term is qq and qth{q^{th}} the term is pp 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 1a,1b,1c,1d\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}.
So, 1q\dfrac{1}{q} is pth{p^{th}} and 1p\dfrac{1}{p} is qth{q^{th}} the term of A.P.
Let the A.P. has aa as the first term and dd as a common difference.
So, the pth{p^{th}} term will be,
⇒1q=a+(p−1)d\Rightarrow \dfrac{1}{q} = a + \left( {p - 1} \right)d ….. (1)
The qth{q^{th}} term will be,
⇒1p=a+(q−1)d\Rightarrow \dfrac{1}{p} = a + \left( {q - 1} \right)d ….. (2)
Subtract equation (2) from equation (1),
⇒1q−1p=a+(p−1)d−a−(q−1)d\Rightarrow \dfrac{1}{q} - \dfrac{1}{p} = a + \left( {p - 1} \right)d - a - \left( {q - 1} \right)d
Simplify the terms,
⇒p−qpq=(p−1−q+1)d\Rightarrow \dfrac{{p - q}}{{pq}} = \left( {p - 1 - q + 1} \right)d
Simplify the terms in the bracket,
⇒(p−q)d=p−qpq\Rightarrow \left( {p - q} \right)d = \dfrac{{p - q}}{{pq}}
Divide both sides by (p−q)\left( {p - q} \right),
⇒d=1pq\Rightarrow d = \dfrac{1}{{pq}}
Substitute the value of dd in equation (1),
⇒1q=a+(p−1)1pq\Rightarrow \dfrac{1}{q} = a + \left( {p - 1} \right)\dfrac{1}{{pq}}
Multiply both sides by pqpq and simplify it,
⇒p=apq+p−1\Rightarrow p = apq + p - 1
Simplify the terms,
⇒1=apq\Rightarrow 1 = apq
Divide both sides by pqpq,
⇒a=1pq\Rightarrow a = \dfrac{1}{{pq}}
Now find the pqthp{q^{th}} term of the AP is,
⇒apq=1pq+(pq−1)1pq\Rightarrow {a_{pq}} = \dfrac{1}{{pq}} + \left( {pq - 1} \right)\dfrac{1}{{pq}}
Take 1pq\dfrac{1}{{pq}} common on the right side,
⇒apq=1pq(1+pq−1)\Rightarrow {a_{pq}} = \dfrac{1}{{pq}}\left( {1 + pq - 1} \right)
Simplify the terms in the bracket,
⇒apq=1pq×pq\Rightarrow {a_{pq}} = \dfrac{1}{{pq}} \times pq
Cancel out the common terms,
⇒apq=1\Rightarrow {a_{pq}} = 1
Thus, the pqthp{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.