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
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