If p^th

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 = p^th term of GP

∴ a = xR^{p – 1}

Similarly, b = xR^{q – 1} & c = xR^{r – 1}

Hence,

(a^{b – c}).(b^{c – a}).(c^{a – b}) = [(xR^{p – 1})^{(q – r)D}].[(xR^{q – 1})^{(r – p)D}].[(xR^{r – 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}

⇒ (a^{b – c}).(b^{c – a}).(c^{a – b}) = x⁰. R⁰

⇒ (a^{b – c}).(b^{c – a}).(c^{a – b}) = 1 …proved