If the $p^{th},q^{th}\:\:and\:\:r^{th}$ terms of G.P.are $a,b\:\:and\:\:c$ respectively then prove that $a^{q-r}.b^{r-p}.c^{p-q}=1$

Toolbox:
• $n^{th}$ term of a G.P.$=a.r^{n-1}$
Given that $p^{th}$ term of a G.P.$=a$
$q^{th}$ term$=b$ and $r^{th}$ term $=c$
Let the first term of the G.P. be $A$ and common ratio =$R$
$\Rightarrow\:t_p=A.R^{p-1}=a$
$t_q=A.R^{q-1}=b$ and
$t_r=A.R^{r-1}=c$
$\Rightarrow\:a^{q-r}=[A.R^{p-1}]^{q-r}=A^{q-r}.R^{(p-1)(q-r)}$
$\Rightarrow\:a^{q-r}=A^{q-r}.R^{(pq-pr-q+r)}$....(i)
Step 2
$b^{r-p}=[A.R^{q-1}]^{r-p}=A^{r-p}.R^{(q-1)(r-p)}$
$\Rightarrow\:b^{r-p}=A^{r-p}.R^{(qr-qp-r+p)}$........(ii)
and
Step 3
$c^{p-q}=[A.R^{r-1}]^{p-q}=A^{p-q}.R^{(r-1)(p-q)}$
$\Rightarrow\:c^{p-q}=A^{p-q}.R^{(rp-rq-p+q)}$........(iii)
Multiplying $(i)\times (ii)\times (iii)$
$\Rightarrow\: a^{q-r}.b^{r-p}.c^{p-q}=$
$[A^{q-r}.R^{(pq-pr-q+r)}]\times [A^{r-p}.R^{(qr-qp-r+p)}]\times[A^{p-q}.R^{(rp-rq-p+q)}]$
$=A^{(q-r+r-p+p-q)}\times R^{(pq-pr-q+r+qr-qp-r+p+rp-rq-p+q)}$
$\Rightarrow\: a^{q-r}.b^{r-p}.c^{p-q}=A^0\times R^0=1\times 1=1$
Hence proved.