If $p$ th, $q$ th and $r$ th terms of an AP and GP are both and $c$ respectively, then show that $a^{b-c} \cdot b^{c-a} \cdot c^{a-b}=1$.
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.
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