If a, b, c are p t h , q t h , and r t h terms of a G.P., then ( c b ) p ( b a ) r ( a c ) q is equal to Options: A . &nbsp 1 B . &nbsp a p b q c r C . &nbsp a q b r c p D . &nbsp a r b p c q

Answer: Option A : A a = A R p − 1 , b = A R q − 1 , c = A R r − 1 ∴ ( c b ) p ( b a ) r ( a c ) q = ( A R p − 1 A R q − 1 ) p ( A R q − 1 A R p − 1 ) r ( A R p − 1 A R r − 1 ) q = R ( r − q ) p + ( q − p ) r + ( p − r ) q = R 0 = 1

If a, b, c are pth, qth, and rth terms of a G.P., then (cb)p(ba)r(ac

Discussion Forum : Sequences And Series

Question -

If a, b, c are $p^{t h}$ , $q^{t h}$ , and $r^{t h}$ terms of a G.P., then $(\frac{c}{b})^{p} (\frac{b}{a})^{r} (\frac{a}{c})^{q}$ is equal to

Options:

A . 1

B .

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

C .

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

D .

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

Answer: Option A
:
A

a = $A R^{p - 1}$ , b = $A R^{q - 1}$ , c = $A R^{r - 1}$

$∴ (\frac{c}{b})^{p} (\frac{b}{a})^{r} (\frac{a}{c})^{q}$ = $(\frac{A R^{p - 1}}{A R^{q - 1}})^{p} (\frac{A R^{q - 1}}{A R^{p - 1}})^{r} (\frac{A R^{p - 1}}{A R^{r - 1}})^{q}$

= $R^{(r - q) p + (q - p) r + (p - r) q}$ = $R^{0}$ = 1

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