Question

If a, b, c are $p^{th}$, $q^{th}$, and $m^{th}$ terms of a G.P., then $\left(\frac{c}{b}{)}^p\right(\frac{b}{a}{)}^m(\frac{a}{c}{)}^q$ 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

Let $A$ be the initial term, and $R$ be the common ratio of the GP.

a = $A{R}^{p-1}$, b = $A{R}^{q-1}$, c = $A{R}^{m-1}$ [$\because T_n=a{r}^{n-1}$]

$\therefore \left(\frac{c}{b}{)}^p\right(\frac{b}{a}{)}^m(\frac{a}{c}{)}^q=(\frac{A{R}^{m-1}}{A{R}^{q-1}}{)}^p\left(\frac{A{R}^{q-1}}{A{R}^{p-1}}{)}^m\right(\frac{A{R}^{p-1}}{A{R}^{m-1}}{)}^q=R^{(m-q)p+(q-p)m+(p-m)q}=R^0=1$