Sol. The result to be proved is based on two basic results of G.P. and H.P.

(i) If p th , q th , r th terms of G.P. are a, b, c then a q–r b r–p c q–p = 1

Proof: We have a = AR p–1 , b = AR q–1 , c = AR r–1 L.H.S. = (AR

p–1 ) q–r (AR q–1 ) r–p (AR r–1 ) p–q = A

q–r +r–p + p –q R (p–1) (q–r)+ (q–1) (r–p) +(r–1)(p–q) = Aº Rº= 1

(ii) If p

th , q th , r th terms of a H.P. are a, b, c then (q–r)bc + (r–c) ca + (p–q) ab = 0

Proof: Let B be the first term, D be the common difference of the corresponding A. P.

We must have B + (p –1) D = 1/a

B + (p –1) D = 1/b …(1)
( ∴ , , are in A.P.)

B + (p –1) D = 1/c

Babc + abc (p –1) = bc ⇒ Babc + abc (q – 1) = ac … (2)

Babc + abc (r –1) = ab From the last system (2) we easily conclude that bc (q – r) + ac (r – p) + ab (p – q) = 0

(i) To prove the given result we make use of the two relations

a q–r b r–p c p–q = 1 … (3)

(q – r) bc + (r – p) ca + (p – q) ab = 0 …(4)

From (3), (q – r)log a + (r – p) log b + (p – q) log c = 0

…(5)

Solving (4) and (5) in the variables q – r, r – p and p – q, we get

=

= = k (say)

⇒ q – r = k(ab log b –ac log c)

r – p = k(bc log c – ab log a)

p – q = k(ca log a – ab log b)

On adding the last three relations, we get

0 = k[–a(b – c) log a –b(c – a) log b – c (a – b) log c]

⇒ a(b – c) log a + b(c – a) log b + c(a – b) log c = 0