Question

if a,b,c are $p^{th},q^{th}and{r}^{th}$ terms of a $GP$, then $\left(\frac{c}{b}{)}^p\right(\frac{b}{a}{)}^r(\frac{a}{c}{)}^q$ is equal to

• A. 1
• B. a^p b^q c^r
• C. a^q b^r c^p
• D. a^r b^p c^q

Answer 1

The correct option is A 1

$a=A{R}^{p-1},b=A{R}^{q-1},c=A{R}^{r-1}$
$\therefore \left(\frac{c}{b}{)}^p\right(\frac{b}{a}{)}^r(\frac{a}{c}{)}^q=(\frac{A{R}^{r-1}}{A{R}^{q-1}}{)}^p\left(\frac{A{R}^{q-1}}{A{R}^{p-1}}{)}^r\right(\frac{A{R}^{p-1}}{ARr-1}{)}^q$
$=R^{(r-q)p+(q-p)r+(p-r)q}=R^{\circ }=1$