Let $A, d$ are the first term and common difference of the AP and $x, R$ are the first term and common ratio of the GP respectively
According to the question
$$
\begin{gathered}
A+(p-1) d=a \\
A+(q-1) d=b \\
A+(r-1) d=c \\
\text { and } a=x R^{p-1} \\
b=x R^{q-1} \\
c=x R^{r-1}
\end{gathered}
$$
On subtracting Eq. (ii) from Eq. (i), we get
$$
\begin{aligned}
d(p-1-q+1) &=a-b \\
\Rightarrow \quad \quad \quad a-b &=d(p-q)
\end{aligned}
$$
On subtracting Eq. (iii) from Eq. (ii), we get
$$
\begin{aligned}
d(q-1-r+1) &=b-c \\
\Rightarrow \quad & b-c &=d(q-r)
\end{aligned}
$$
On subtracting Eq. (i) form Eq. (iii), we get
$$
\begin{aligned}
&\mathrm{d}(\mathrm{r}-1-\mathrm{p}+1)=\mathrm{c}-\mathrm{a} \\
&\Rightarrow \mathrm{c}-\mathrm{a}=\mathrm{d}(\mathrm{r}-\mathrm{p}) \\
&\mathrm{Now},=a^{b-c} b^{c-a} c^{a-b} \\
&=\left(x R^{p-1}\right)^{d(q-r)}\left(x R^{q-1}\right)^{d(r-p)}\left(x R^{r-1}\right) d(p-q)
\end{aligned}
$$
Using Eqns. (iv), (v), (vi) and (vii), (viii), (ix),
$=x^{d(q-r)+d(r-p)+d(p-q)} R^{(p-1) d(q-r)}$
$+(q-1) d(r-p)+(r-1) d(p-q)$
$=x^{d(q-r+r-p+p-q)}$
$R^{d(p q-p r-q+r+q r-p q-r+p+r p-r q-p+q)}$
$=x^0 R^0=1=\mathrm{RHS}$
Hence proved.