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Mathematics Question on geometric progression

If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq−rbr−pcp−q=1.a^{q-r} b^{r-p} c^{p-q}=1.

Answer

Let A be the first term and R be the common ratio of the G.P.
According to the given information,
ARp-1 = a
ARq-1 = b
ARr-1 = c
aq−rbr−pcp−qa^{q-r}b^{ r-p} c^{ p-q}
=Aq−r×R(p−1)(q−r)cAr−p×R(q−1)(r−p)×Ap−q×R(r−1)(p−q)= A^{ q-r} ×R^{(p-1)(q-r)}c A ^{r-p} × R ^{(q-1) (r-p)} × A^{ p-q }× R ^{(r -1)(p-q)}
=Aq−r+r−p+p−q×R(pr−pr−q+r)+(rq−r+p−pq)+(pr−p−qr+q)= Aq^{ - r + r - p + p - q }× R ^{(pr- pr - q+ r) + (rq- r + p- pq) + (pr- p - qr+ q)}
= A0 x R0
= 1
Thus, the given result is proved.