If $p^{th},q^{th}$ and $r^{th}$ term of a G.P. are $x,y$ and $z$ respectively prove that :
$x^{a-r}x{y}^{r-p}x{z}^{p-q}=1$.

Given,

$a{r}^{p-1}=x$

$a{r}^{q-1}=y$

$a{r}^{r-1}=z$

$\Rightarrow (a{r}^{p-1}{)}^{q-r}=x^{q-r}$

$\Rightarrow (a{r}^{q-1}{)}^{r-p}=y^{r-p}$

$\Rightarrow (a{r}^{r-1}{)}^{p-q}=z^{p-q}$

$x^{q-r}{y}^{r-p}{z}^{p-q}=(a{r}^{p-1}{)}^{q-r}(a{r}^{q-1}{)}^{r-p}(a{r}^{r-1}{)}^{p-q}$

$x^{q-r}x{y}^{r-p}x{z}^{p-q}=a^{(p+q+r-p-q-r)}{r}^{(pq+qr+rp+p+q+r-qp-rq-pr-p-q-r)}$

$\Rightarrow x^{q-r}x{y}^{r-p}x{z}^{p-q}=a^0{r}^0$

$\therefore x^{q-r}x{y}^{r-p}x{z}^{p-q}=1$

Hence proved.