Let first term and common difference of the AP be a and $d$, respectively.

$$\begin{aligned} \text{Then,}\quad S_p & =q \\ \Rightarrow \quad \frac{p}{2}[2 a+(p-1) d] & =q \\ 2 a+(p-1) d & =\frac{2 q}{p} \quad \text{... (i)}\\ \text{and}\quad S_q & =p \\ \frac{q}{2}[2 a+(q-1) d] & =p \\ 2 a+(q-1) d & =\frac{2 p}{q}\quad \text{... (ii)} \end{aligned}$$

On subtracting Eq. (ii) from Eq. (i), we get

$$\begin{aligned} & 2 a+(p-1) d-2 a-(q-1) d=\frac{2 q}{p}-\frac{2 p}{q} \\ & \Rightarrow \quad[(p-1)-(q-1)] d=\frac{2 q^2-2 p^2}{p q} \\ & \Rightarrow \quad[p-1-q+1] d=\frac{2\left(q^2-p^2\right)}{p q} \\ & \Rightarrow \quad(p-q) d=\frac{2\left(q^2-p^2\right)}{p q} \\ & \therefore \quad d=\frac{-2(p+q)}{p q}\quad \text{.... (iii)} \end{aligned}$$

$$\begin{aligned} &\text { On substituting the value of } d \text { in Eq. (i), we get }\\ &\begin{aligned} 2 a+(p-1)\left(\frac{-2(p+q)}{p q}\right) & =\frac{2 q}{p} \\ 2 a & =\frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q} \\ a & =\left[\frac{q}{p}+\frac{(p+q)(p-1)}{p q}\right]\quad \text{.... (iv)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Now, }\quad S_{p+q} & =\frac{p+q}{2}[2 a+(p+q-1) d] \\ & =\frac{p+q}{2}\left[\frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q}-\frac{(p+q-1) 2(p+q)}{p q}\right] \\ & =(p+q)\left[\frac{q}{p}+\frac{(p+q)(p-1)-(p+q-1)(p+q)}{p q}\right] \\ & =(p+q)\left[\frac{q}{p}+\frac{(p+q)(p-1-p-q+1)}{p q}\right] \\ & =p+q\left[\frac{q}{p}-\frac{p+q}{p}\right]=(p+q)\left[\frac{q-p-q}{p}\right] \\ S_{p+q} & =-(p+q) \\ S_{p-q} & =\frac{p-q}{2}[2 a+(p-q-1) d] \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{p-q}{2}\left[\frac{2 q}{p}+\frac{2(p+q)(p-1)}{p q}-\frac{(p-q-1) 2(p+q)}{p q}\right] \\ & =(p-q)\left[\frac{q}{p}+\frac{p+q(p-1-p+q+1)}{p q}\right] \\ & =(p-q)\left[\frac{q}{p}+\frac{(p+q) q}{p q}\right] \\ & =(p-q)\left[\frac{q}{p}+\frac{p+q}{p}\right]=(p-q) \frac{(p+2 q)}{p} \end{aligned}$$

Let $A, d$ are the first term and common difference of $A P$ and $x, R$ are the first term and common ratio of GP, respectively.

According to the given condition,

$$\begin{aligned} A+(p-1) d & =a \quad \text{... (i)}\\ A+(q-1) d & =b \quad \text{... (ii)}\\ A+(r-1) d & =c \quad \text{... (iii)}\\ \text{and}\quad a & =x R^{p-1} \quad \text{... (iv)}\\ b & =x R^{q-1} \quad \text{... (v)}\\ c & =x R^{r-1} \quad \text{... (vi)} \end{aligned}$$

On subtracting Eq. (ii) from Eq. (i), we get

$$\begin{aligned} d(p-1-q+1) & =a-b \\ a-b & =d(p-q)\quad \text{... (vii)} \end{aligned}$$

On subtracting Eq. (iii) from Eq. (ii), we get

$$\begin{aligned} d(q-1-r+1) & =b-c \\ b-c & =d(q-r)\quad \text{... (viii)} \end{aligned}$$

On subtracting Eq. (i) from Eq. (iii), we get

$$\begin{aligned} d(r-1-p+1) & =c-a \\ c-a & =d(r-p)\quad \text{... (ix)} \end{aligned}$$

$\text { Now, we have to prove } a^{b-c} b^{c-a} c^{a-b}=1$

$$\text { Taking LHS }=a^{b-c} b^{c-a} c^{a-b}$$

Using Eqs. (iv), (v), (vi) and (vii), (viii), (ix),

$$\begin{aligned} \text { LHS } & =\left(x R^{p-1}\right)^{d(q-r)}\left(x R^{q-1}\right)^{d(r-p)}\left(x R^{r-1}\right)^{d(p-q)} \\ & =x^{d(q-r)+d(r-p)+d(p-q)} R^{(p-1) d(q-r)+(q-1) d(r-p)+(r-1) d(p-q)} \\ & =x^{d(q-r+r-p+p-q)} \end{aligned}$$

$$\begin{aligned} R^{d(p q-p r-q+r+q r-p q-r+p+r p-r q-p+q)} & =x^0 R^0=1 \\ & =\text { RHS } \end{aligned}$$

Hence proved.