Let the numbers be $a$ and $b$.

Then,

$$\begin{aligned} A & =\frac{a+b}{2} \\ \Rightarrow \quad 2 A & =a+b\quad \text{.... (i)} \end{aligned}$$

and $G_1, G_2$ be geometric mean between $a$ and $b$, then $a, G_1, G_2, b$ are in GP.

Let $r$ be the common ratio.

Then,

$$\begin{aligned} & b=a r^{4-1} \quad \left[\because a_n=a r^{n-1}\right] \\ \Rightarrow\quad& b=a r^3 \Rightarrow \frac{b}{a}=r^3 \\ \therefore \quad& r=\left(\frac{b}{a}\right)^{1 / 3} \end{aligned}$$

Now, $$G_1=a r=a\left(\frac{b}{a}\right)^{1 / 3} \quad\left[\because r=\left(\frac{b}{a}\right)^{1 / 3}\right]$$

and $$G_2=a r^2=a\left(\frac{b}{a}\right)^{2 / 3}$$

$$\begin{aligned} \text{RHS} & =\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=\frac{\left[a\left(\frac{b}{a}\right)^{1 / 3}\right]^2}{a\left(\frac{b}{a}\right)^{2 / 3}}+\frac{\left[a\left(\frac{b}{a}\right)^{2 / 3}\right]^2}{a\left(\frac{b}{a}\right)^{1 / 3}} \\ & =\frac{a^2\left(\frac{b}{a}\right)^{2 / 3}}{a\left(\frac{b}{a}\right)^{2 / 3}}+\frac{a^2\left(\frac{b}{a}\right)^{4 / 3}}{a\left(\frac{b}{a}\right)^{1 / 3}} \\ & =a+a\left(\frac{b}{a}\right)=a+b=2 A \quad \text{[using Eq. (i)]}\\ & =\text { LHS } \end{aligned}$$

Since, $\theta_1, \theta_2, \theta_3, \ldots, \theta_n$ are in AP.

$$\Rightarrow \quad \theta_2-\theta_1=\theta_3-\theta_2=\cdots=\theta_n-\theta_{n-1}=d\quad \text{.... (i)}$$

Now, we have to prove

$$\sec q_1 \sec q_2+\sec q_2 \sec q_3+\cdots+\sec q_{n-1} \sec \theta_n=\frac{\tan \theta_n-\tan \theta_1}{\sin d}$$

or it can be written as

$$\sin d\left[\sec \theta_1 \sec \theta_2+\sec \theta_2 \sec \theta_3+\cdots+\sec \theta_{n-1} \sec \theta_n\right]=\tan \theta_n-\tan \theta_1$$

Now, taking only first term of LHS

$$\begin{aligned} &\begin{aligned} \sin d \sec \theta_1 \sec \theta_2 & =\frac{\sin d}{\cos \theta_1 \cos \theta_2}=\frac{\sin \left(\theta_2-\theta_1\right)}{\cos \theta_1 \cos \theta_2} \quad \text{\text { [from Eq. (i)] }}\\ & =\frac{\sin \theta_2 \cos \theta_1-\cos \theta_2 \sin \theta_1}{\cos \theta_1 \cos \theta_2} \\ & \quad[\because \sin (A-B)=\sin A \cdot \cos B-\cos A \cdot \sin B] \\ & =\frac{\sin \theta_2 \cos \theta_1}{\cos \theta_1 \cos \theta_2}-\frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1 \cos \theta_2}=\tan \theta_2-\tan \theta_1 \end{aligned}\\ \end{aligned}$$

Similarly, we can solve other terms which will be $\tan \theta_3-\tan \theta_2, \tan \theta_4-\tan \theta_3, \cdots$

$$\begin{aligned} \therefore \quad \mathrm{LHS} & =\tan \theta_2-\tan \theta_1+\tan \theta_3-\tan \theta_2+\cdots+\tan \theta_n-\tan \theta_{n-1} \\ & =-\tan \theta_1+\tan \theta_n=\tan \theta_n-\tan \theta_1 \quad \text { Hence proved. } \\ & =\text { RHS } \quad \text { Hence proved. } \end{aligned}$$

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.