If a- b- c are p-th- nbsp-q-th- and nbsp-r-th terms of a G-P- then -c-b-p-b-a-r-a-c-q is equal to -1-

The correct option is D 1 a = AR^p 1, b = AR^q 1, c = AR^r 1 ∴ (c/b)^p (b/a)^r(a/c)^q = (AR^p 1/AR^q 1)^p (AR^q 1/AR^p 1)^r nbsp;(AR^p 1/AR^r 1)^q = R^(r q)p ...

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Byju's Answer

Standard XI

Mathematics

nth Term of GP

If a, b, c ar...

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

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Solution

The correct option is A
1

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

The correct option is A 1 a = ARp−1, b = ARq−1, c = ARr−1 ∴(cb)p(ba)r(ac)q = (ARp−1ARq−1)p(ARq−1ARp−1)r(ARp−1ARr−1)q = R(r−q)p+(q−p)r+(p−r)q = R0 = 1

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