Question

The $ {p^{th}} $ term of H.P. is $ q(p + q) $ and \[{q^{th}}\] term $ {T_q} $ is $ p(p + q) $ where $ p > 1 $ , $ q > 1 $ where $ p \ne q $ , then
A. $ {T_{p + q}} = pq $
B. $ {T_{pq}} = p + q $
C. $ {T_{p + q}} > {T_{pq}} $
D. $ {T_{pq}} > {T_{p + q}} $

Answer 1

Hint: In this question, we are given the value of two unknown terms of an H.P. and we have to find the value of some other terms using the given values. It can be done by using the formula for the nth term of the H.P., from that formula we will find the value of the first term and the common difference of the H.P. to find the value of required terms.

Complete step by step solution:
The formula for finding any term of an H.P. is $ {T_n} = \dfrac{1}{{a + (n - 1)d}} $
 $ {T_p} = \dfrac{1}{{a + (p - 1)d}} $ but we have $ {T_p} = q(p + q) $
So, $ \dfrac{1}{{a + (p - 1)d}} = q(p + q) $ or $ a + (p - 1)d = \dfrac{1}{{q(p + q)}}...(1) $
 $ {T_q} = \dfrac{1}{{a + (q - 1)d}} $ but we have $ {T_q} = p(p + q) $
So, $ \dfrac{1}{{a + (q - 1)d}} = p(p + q) $ or $ a + (q - 1)d = \dfrac{1}{{p(p + q)}}...(2) $
Now $ (1) - (2) $ -
 $
  a + (p - 1)d - [a + (q - 1)d] = \dfrac{1}{{q(p + q)}} - \dfrac{1}{{p(p + q)}} \\
   \Rightarrow a + (p - 1)d - a - (q - 1)d = \dfrac{1}{{p + q}}(\dfrac{{p - q}}{{pq}}) \\
   \Rightarrow d(p - q) = \dfrac{{p - q}}{{pq(p + q)}} \\
   \Rightarrow d = \dfrac{1}{{pq(p + q)}} \\
  $
Putting the value of d in (1), we get –
 $
  a + (p - 1)\dfrac{1}{{pq(p + q)}} = \dfrac{1}{{q(p + q)}} \\
   \Rightarrow a = \dfrac{1}{{q(p + q)}} - \dfrac{{p - 1}}{{pq(p + q)}} \\
   \Rightarrow a = \dfrac{{p - p + 1}}{{pq(p + q)}} \\
   \Rightarrow a = \dfrac{1}{{pq(p + q)}} \\
  $
We get $ a = d $ , so for this H.P., we get –
 $
  {T_n} = \dfrac{1}{{a + (n - 1)d}} = \dfrac{1}{{a + (n - 1)a}} = \dfrac{1}{{a + an - a}} \\
   \Rightarrow {T_n} = \dfrac{1}{{an}} \\
  $
So,
 $
  {T_{p + q}} = \dfrac{1}{{\dfrac{1}{{pq(p + q)}}(p + q)}} \\
   \Rightarrow {T_{pq}} = \dfrac{1}{{\dfrac{1}{{pq}}}} = pq \\
  $
Thus, option (A) is correct.
 $
  {T_{pq}} = \dfrac{1}{{\dfrac{1}{{pq(p + q)}}pq}} \\
   \Rightarrow {T_{pq}} = \dfrac{1}{{\dfrac{1}{{p + q}}}} = p + q \;
  $
Thus, option (B) is also correct.
Now, we are given that $ p > 1,\,q > 1\,and\,p \ne q $ , let $ p = 3 $ and $ q = 4 $ , so –
 $ {T_{p + q}} = pq = 3(4) = 12 $ and $ {T_{pq}} = p + q = 3 + 4 = 7 $
Clearly $ {T_{p + q}} > {T_{pq}} $ .
Hence option (A), (B) and (C) are correct.

Note: There are various kinds of sequences in mathematics; one of them is a Harmonic progression. A harmonic progression is the reciprocal of arithmetic progression. In an arithmetic progression, each term is equal to the sum of the previous term and the common difference. It is of the form $ a,\,a + d,\,a + 2d.... $ . So, harmonic progression is of the form $ \dfrac{1}{a},\,\dfrac{1}{{a + d}},\,\dfrac{1}{{a + 2d}}... $ . This way, the nth term of an H.P. is $ \dfrac{1}{{a + (n - 1)d}} $ . we can also say that each term of the H.P. is the harmonic mean of its neighboring terms.