If a, b, c are the p^th, q^th and r^th terms of a GP, show that

(q – r) log a + (r – p) log b + (p – q) log c = 0.

As per the question, a, b and c are the p^th, q^th and r^th term of GP.

Let us assume the required GP as A, AR, AR², AR³…

Now, the n^th term in the GP, a_n = AR^n-1

p^th term, a_p = AR^p-1 = a → (1)

q^th term, a_q = AR^q-1 = b → (2)

r^th term, a_r = AR^r-1 = c → (3)

→ (i)

→ (ii)

→ (iii)

Taking logarithm on both sides of equation (i), (ii) and (iii).

(p – q) log R = log a – log b,

∴ → (4)

(q – r) log R = log b – log c

∴ → (5)

(r – p) log R = log c – log a

∴ → (6)

Now, multiply equation (4) with log c,

→ (7)

Now, multiply equation (5) with log a,

→ (8)

Now, multiply equation (6) with log b,

→ (9)

Now, add equations (7), (8) and (9).

On solving the above equation, we will get,

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

Hence proved.