# If pth

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

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