If pth, qth and rth terms of an A.P. and G.P. are both a, b and c respectively, show that ab – cbc – a ca – b = 1.
Let the A.P. be A, A + D, A + 2 D, ... and G.P be x, xR, xR2, ... then
a = A + (p – 1)D, b = A + (q – 1)D, c = A + (r – 1)D
⇒ a – b = (p – q)D
Also, b – c = (q – r)D
And, c – a = (r – p)D
Also a = pth term of GP
∴ a = xRp – 1
Similarly, b = xRq – 1 & c = xRr – 1
Hence,
(ab – c).(bc – a).(ca – b) = [(xRp – 1)(q – r)D].[(xRq – 1)(r – p)D].[(xRr – 1)(p – q)D]
= x(q – r + r – p + p – q)D. R[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]D
⇒ (ab – c).(bc – a).(ca – b) = x0. R0
⇒ (ab – c).(bc – a).(ca – b) = 1 …proved