If the #p^(th), q^(th), and r^(th)# of a H.P is a,b,c respectively, then prove that#(q-r) /a + (r-p) / b + (p-q)/c = 0#?

1 Answer
Shwetank Mauria
Aug 21, 2017

Please see below.

Explanation:

Before we commence with the question, two things first.

  1. If #a_m# and #a_n# are #m^(th)# and #n^(th)# terms of an Arithmetic Progression, whose common difference is #d#, then #a_m-a_n=(m-n)d#.
  2. If #h_1,h_2,h_3,h_4,.....# are in H.P., then #1/h_1,1/h_2,1/h_3,1/h_4,.....# are in A.P.

Hence, as #p^(th)#, #q^(th)# and #r^(th)# terms of a H.P. are #a#, #b# and #c#, then #1/a#, #1/b# and #1/c# are #p^(th)#, #q^(th)# and #r^(th)# terms of an A.P. and hence

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

or #p-q=1/d(1/a-1/b)# ......................(1)

Similarly #r-p=1/d(1/c-1/a)# ......................(2)

and #q-r=1/d(1/b-1/c)# ......................(3)

Hence #(q-r)/a+(r-p)/b+(p-q)/c#

= #1/d[(1/b-1/c)1/a+(1/c-1/a)1/b+(1/a-1/b)1/c]#

= #1/d[1/(ab)-1/(ca)+1/(bc)-1/(ab)+1/(ca)-1/(bc)]#

= #1/d xx0=0#

Related questions
Impact of this question
1935 views around the world
You can reuse this answer
Creative Commons License