Solution:

Let m be the first term of the AP and d be the common difference

Let I be the GP’s first term and s be the common ration

The Ap’s nth term is given as $t_{n}=a+{(n-1)}d$ in which the first
term is a and the common difference is d

The GP’s nth term is given by $t_{n}={{ar}^{n-1}}$ in which the first term is a snd the common ratio is r.

Both AP and GP’s pth term is a

For the AP

$\Rightarrow {t_{p}}={m+(p-1)d}$

$\Rightarrow {a=m+(p-1)d….1}$

For the GP

$\Rightarrow {t_{p}}={{Is}^{p-1}}$

$\Rightarrow {a={{Is}^{p-1}}……2}$

Both AP and GP’s qth term is b

For the AP

$\Rightarrow {{t_{q}}={m+(q-1)d}}$

$\Rightarrow {b={m+(q-1)d}…….3}$

For the GP

$\Rightarrow \mathrm{t}_{\mathrm{q}}=\mathrm{Is}^{\mathrm{q}-1}$

$\Rightarrow \mathrm{b}=\mathrm{Is}^{\mathrm{q}-1} \ldots . . .4$ The

$\mathrm{r}^{\text {th }}$ term $\left(\mathrm{t}_{\mathrm{r}}\right)$
of both $\mathrm{AP}$ and $\mathrm{GP}$ is $\mathrm{c}$

For the AP

$\begin{array}{l}
\Rightarrow \mathrm{t}_{r}=m+(r-1) \mathrm{d} \\
\Rightarrow c=m+(r-1) d \ldots . .5
\end{array}$

For the GP

$\begin{array}{l}
\Rightarrow \mathrm{t}_{r}=\mathrm{Is}^{\mathrm{r-1}} \\
\Rightarrow \mathrm{c}=1 \mathrm{~s}^{r-1} \ldots \ldots 6
\end{array}$

Let’s now find $\mathbf{b}-\mathrm{c}, \mathrm{c}-\mathrm{a}$ and $\mathrm{a}-\mathrm{b}$

Using the eq.3 and eq.5

$\Rightarrow \mathrm{b}-\mathrm{c}=(\mathrm{q}-\mathbf{r}) \mathrm{d} \ldots(\mathrm{i})$

$\Rightarrow c-a=(r-p) d \ldots(i i)$

Using the eq.1 and eq.3

$\Rightarrow a-b=(p-q) d \ldots \text { (iii) }$

We need to prove that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$

$\left.\mathrm{LHS}=\mathrm{a}^{b-c} \cdot \mathrm{b}^{c-a} \cdot c^{a-b}\right)$

Using the eq.2, eq.4 and eq.6

$\Rightarrow$ Left Hand Side $=\left(I s^{p-1}\right)^{b-c} \cdot\left(I s^{q-1}\right)^{c-a} \cdot\left(I s^{r-1}\right)^{a-b}$

$=\left(\frac{l s^{p}}{s}\right)^{b-c} \cdot\left(\frac{l s^{q}}{s}\right)^{c-a} \cdot\left(\frac{l s^{r}}{s}\right)^{a-b}$

$=\frac{1^{b-c} s^{p(b-c)}}{s^{b-c}} \cdot \frac{1^{c-a} s^{q(c-a)}}{s^{c-a}} \cdot \frac{1^{a-b} s^{r(a-b)}}{s^{a-b}}$

$=\frac{1^{b-c+c-a+a-b}}{s^{b-c+c-a+a-b} \cdot s^{p(b-c)} \cdot s^{q(c-a)} \cdot s^{r(a-b)}}$

${=} {s^{p(b-c)} \cdot s^{q(c-a)} \cdot s^{r(a-b)}}$

Now substituting the values of $a-b, c-a$ and $b-c$ from eq.(iii), eq.(ii) and eq.(i)

$={s^{p(q-r)d}}.{s^{q(r-p)d}}.{s^{r(p-q)d}}$

$={s^{{pqd}-{prd}}}.{s^{{qrd}-{pqd}}}.{s^{{prd}-{qrd}}}$

$={{s}^{{pqd}-{prd}+{qrd}-{pqd}+{prd}-{qrd}}}$

$={s}^{0}=1$

$\Rightarrow$ Left Hand Side=Right Hand Side

As a result, hence proved.