If the pth, qth and rth terms of a GP are a, b, c, then ${(\frac{c}{b})}^{p} {(\frac{b}{a})}^{r} {(\frac{a}{c})}^{q}$ is equal to

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$a^{'} b^{p} c^{q}$

$a^{q} b^{r} c^{p}$

$a^{p} b^{q} c^{r}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given, pth term= a, qth term= b, rth term= c

Let A be the first term and R be the common ratio of the GP

$\begin{aligned} ∴ A R^{ρ - 1} = a \\ A R^{a - 1} = b \\ A R^{r - 1} = c \end{aligned}$

On dividing Eq. (iii) by Eq. (ii), we get

$\frac{c}{b} = R^{r - a}$

On dividing Eq. (ii) by Eq. (i), we get

$\frac{b}{a} = R^{q - ρ}$

On dividing Eq. (i) by Eq. (iii), we get

$\frac{b}{a} = R^{q - p}$

Now, $\begin{aligned} {(\frac{c}{b})}^{ρ} \cdot {(\frac{b}{a})}^{r} \cdot {(\frac{a}{c})}^{q} = {(R^{r - a})}^{p} \cdot {(R^{q - p})}^{r} \cdot (R^{ρ - 1} p \\ = R^{p r - ρ q + r q - r p + q p - q r} \\ = R^{0} = 1 [from Eqs. (iv), (v) and (vi)] \end{aligned}$