If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an A.P. and G.P. are both $a, b$ and $c$ respectively, show that $a^{b-c} b^{c-a} c^{a-b}=1$.
Let A be the first term and D be the common difference of the AP. Therefore,
$a_{p}=A+(p-1) D=a \quad \ldots(1)$
$a_{q}=A+(q-1) D=b \quad \ldots(2)$
$a_{r}=A+(r-1) D=c \quad \ldots(3)$
Also, suppose A' be the first term and R be the common ratio of the GP. Therefore,
$a_{p}=A^{\prime} R^{p-1}=a \quad \ldots \ldots(4)$
$a_{q}=A^{\prime} R^{q-1}=b \quad \ldots(5)$
$a_{r}=A^{\prime} R^{r-1}=c \quad \ldots \ldots(6)$
Now,
Subtracting (2) from (1), we get
$A+(p-1) D-A-(q-1) D=a-b$
$\Rightarrow(p-q) D=a-b \quad \ldots(7)$
$A+(q-1) D-A-(r-1) D=b-c$
$\Rightarrow(q-r) D=b-c \quad \ldots(8)$
Subtracting (1) from (3), we get
$A+(r-1) D-A-(p-1) D=c-a$
$\Rightarrow(r-p) D=c-a \quad \ldots(9)$
$\therefore a^{b-c} b^{c-a} c^{a-b}$
$=\left[A^{\prime} R^{(p-1)}\right]^{(q-r) D} \times\left[A^{\prime} R^{(q-1)}\right]^{(r-p) D} \times\left[A^{\prime} R^{(r-1)}\right]^{(p-q) D} \quad[$ Using $(4),(5),(6),(7),(8)$ and $(9)]$
$=A^{,(q-r) D} R^{(p-1)(q-r) D} \times A^{,(r-p) D} R^{(q-1)(r-p) D} \times A^{,(p-q) D} R^{(r-1)(p-q) D}$
$=A^{,[(q-r) D+(r-p) D+(p-q) D]} \times R^{[(p-1)(q-r) D+(q-1)(r-p) D+(r-1)(p-q) D]}$
$=A^{\prime}[q-r+r-p+p-q] D \times R^{[p q-p r-q+r+q r-p q-r+p+p r-q r-p+q] D}$
$=\left(A^{\prime}\right)^{0} \times R^{0}$
$=1 \times 1$
$=1$