Question

In a Harmonic Progression, pth term is q and the qth term is p. Then pqth term is

• A. 0
• B. 1
• C. pq
• D. pq(p+q)

Hint:

A problem involving a Harmonic Progression (HP) should always be converted into a problem about an Arithmetic Progression (AP) by taking the reciprocals of the terms. This standard result for HP (if p-th term is q and q-th term is p, then pq-th term is 1) is a direct parallel to a similar result in AP.

Answer 1

The Correct Option is B

If a sequence is a Harmonic Progression (HP), then the sequence of its reciprocals is an Arithmetic Progression (AP). 

Let the HP be \( h_1, h_2, \ldots \). The corresponding AP is \( a_1, a_2, \ldots \) where \( a_n = 1/h_n \). 

Let the first term of the AP be \( A \) and the common difference be \( D \). 

The n-th term of the AP is \( a_n = A + (n-1)D \). 

We are given: \( p^{th} \) term of HP is \( q \implies h_p = q \implies a_p = 1/q \). \( q^{th} \) term of HP is \( p \implies h_q = p \implies a_q = 1/p \). 

Using the AP formula: \[ A + (p-1)D = 1/q \cdots(1) \] \[ A + (q-1)D = 1/p \cdots(2) \] Subtract (2) from (1): \[ (p-1)D - (q-1)D = \frac{1}{q} - \frac{1}{p} \] \[ (p-q)D = \frac{p-q}{pq} \implies D = \frac{1}{pq} \] Substitute D back into (1): \[ A + (p-1)\frac{1}{pq} = \frac{1}{q} \implies A = \frac{1}{q} - \frac{p-1}{pq} = \frac{p - (p-1)}{pq} = \frac{1}{pq} \] So, for the AP, \( A = 1/pq \) and \( D = 1/pq \). We need to find the \( pq^{th} \) term of the HP, which is \( h_{pq} \). First, find the \( pq^{th} \) term of the AP, \( a_{pq} \). \[ a_{pq} = A + (pq-1)D = \frac{1}{pq} + (pq-1)\frac{1}{pq} = \frac{1 + pq - 1}{pq} = \frac{pq}{pq} = 1 \] The \( pq^{th} \) term of the HP is the reciprocal of the \( pq^{th} \) term of the AP. \[ h_{pq} = \frac{1}{a_{pq}} = \frac{1}{1} = 1 \]