If a, b, c are the pth, qth and rth 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 pth, qth and rth term of GP.
Let us assume the required GP as A, AR, AR2, AR3…
Now, the nth term in the GP, an = ARn-1
pth term, ap = ARp-1 = a → (1)
qth term, aq = ARq-1 = b → (2)
rth term, ar = ARr-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.