Question

If p^th, q^th and r^th terms of an A.P. and G.P. are both a, b and c respectively, show that $a^{b-c}{b}^{c-a}{c}^{a-b}=1$.

Answer 1

Let A be the first term and D be the common difference of the AP. Therefore,

$$\begin{gathered} a_p=A+\left(p-1\right)D=a.....\left(1\right) \\ {a}_q=A+\left(q-1\right)D=b.....\left(2\right) \\ {a}_r=A+\left(r-1\right)D=c.....\left(3\right) \end{gathered}$$

Also, suppose A' be the first term and R be the common ratio of the GP. Therefore,

$$\begin{gathered} a_p=A\text{'}{R}^{p-1}=a.....\left(4\right) \\ {a}_q=A\text{'}{R}^{q-1}=b.....\left(5\right) \\ {a}_r=A\text{'}{R}^{r-1}=c.....\left(6\right) \end{gathered}$$

Now,

Subtracting (2) from (1), we get

$$\begin{gathered} A+\left(p-1\right)D-A-\left(q-1\right)D=a-b \\ \Rightarrow \left(p-q\right)D=a-b.....\left(7\right) \end{gathered}$$

Subtracting (3) from (2), we get

$$\begin{gathered} A+\left(q-1\right)D-A-\left(r-1\right)D=b-c \\ \Rightarrow \left(q-r\right)D=b-c.....\left(8\right) \end{gathered}$$

Subtracting (1) from (3), we get

$$\begin{gathered} A+\left(r-1\right)D-A-\left(p-1\right)D=c-a \\ \Rightarrow \left(r-p\right)D=c-a.....\left(9\right) \end{gathered}$$

$\therefore a^{b-c}{b}^{c-a}{c}^{a-b}$

$={\left[A\text{'}{R}^{\left(p-1\right)}\right]}^{\left(q-r\right)D}\times {\left[A\text{'}{R}^{\left(q-1\right)}\right]}^{\left(r-p\right)D}\times {\left[A\text{'}{R}^{\left(r-1\right)}\right]}^{\left(p-q\right)D}$ [Using (4), (5), (6), (7), (8) and (9)]

$=A{\text{'}}^{\left(q-r\right)D}{R}^{\left(p-1\right)\left(q-r\right)D}\times A{\text{'}}^{\left(r-p\right)D}{R}^{\left(q-1\right)\left(r-p\right)D}\times A{\text{'}}^{\left(p-q\right)D}{R}^{\left(r-1\right)\left(p-q\right)D}$

$=A{\text{'}}^{\left[\left(q-r\right)D+\left(r-p\right)D+\left(p-q\right)D\right]}\times R^{\left[\left(p-1\right)\left(q-r\right)D+\left(q-1\right)\left(r-p\right)D+\left(r-1\right)\left(p-q\right)D\right]}$

$$\begin{gathered} =A{\text{'}}^{\left[q-r+r-p+p-q\right]D}\times R^{\left[pq-pr-q+r+qr-pq-r+p+pr-qr-p+q\right]D} \\ ={\left(A\text{'}\right)}^0\times R^0 \\ =1\times 1 \\ =1 \end{gathered}$$