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 # so True
Part (B)
# T_(pq) = (pq(p+q))/(pq) = p+q # so True
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 # so False
(D)# T_(pq) gt T_(p+q) => p+q gt pq # so True
