# The #p^th# term #T_p# of H.P (Harmonic Progression) is #q(p+q)# and #q^th# term #T_q# is #p(p+q)# when p>1, q>1 then? The question has multiple answers.

##
A) #T_(p+q)# = #pq#

B) #T_(pq) = p + q#

C) #T_(p+q) > T_(pq)#

D) #T_(pq) > T_(p+q)#

A)

B)

C)

D)

##### 1 Answer

(A) = True

(B) = True

(C) = False

(D) = True

#### Explanation:

A Harmonic Progression is formed by taking the reciprocals of the terms of an Arithmetic Progression. Thus we have a AP sequence:

# {a, a+d, a+2d, ..., a+(n-1)d }#

The corresponding Harmonic Progression sequence is:

# { 1/a, 1/(a+d), 1/(a+2d), ..., 1/(a+(n-1)d) }#

So we can write

# T_p = 1/(a+(p-1)d) = q(p+q) #

# T_q = 1/(a+(q-1)d) = p(p+q) #

Inverting each equation we get:

# a+(p-1)d = 1/(q(p+q)) # ..... [A]

# a+(q-1)d = 1/(p(p+q)) # ..... [B]

Then Eq[A]-Eq[B] we get:

# a+(p-1)d - (a+(q-1)d) = 1/(q(p+q)) - 1/(p(p+q)) #

# :. a+(p-1)d - a-(q-1)d = (p-q)/(pq(p+q)) #

# :. d{(p-1) - (q-1)} = (p-q)/(pq(p+q)) #

# :. d(p-1 - q+1) = (p-q)/(pq(p+q)) #

# :. d(p - q) = (p-q)/(pq(p+q)) #

# :. d = 1/(pq(p+q)) \ \ \ \ # provided#p != q#

Substituting this result into [A] we get:

# a+(p-1)/(pq(p+q)) = 1/(q(p+q)) #

# :. a = p/(pq(p+q)) - (p-1)/(pq(p+q))#

# :. a = (p-(p-1))/(pq(p+q))#

# :. a = 1/(pq(p+q))#

Thus we have

# u_n = a+(n-1)d #

# \ \ \ \ = 1/(pq(p+q)) + (n-1)/(pq(p+q)) #

# \ \ \ \ = (1+(n-1))/(pq(p+q))#

# \ \ \ \ = n/(pq(p+q))#

And so we have, the

# T_n = 1/(a+(n-1)d) #

# \ \ \ \ = (pq(p+q))/n #

Now we have all information required to answer the question

**Part (A)**

# T_(p+q) = (pq(p+q))/(p+q) = pq # soTrue

**Part (B)**

# T_(pq) = (pq(p+q))/(pq) = p+q # soTrue

**Part (C) & (D)**

We have

# p gt 1 => pq gt q #

# q gt 1 => pq gt p #

Adding these results we get:

# pq + pq gt p + q #

# :. 2 pq > p+q #

# :. pq > 1/2(p+q) lt p+q #

Then using the results from (A) and (B):

(C)

# T_(p+q) gt T_(pq) => pq gt p+q # soFalse

(D)# T_(pq) gt T_(p+q) => p+q gt pq # soTrue