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

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8. Sequences and Series

easy

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

$1$

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

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

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

Solution

(a) $a = A{R^{p – 1}},\;b = A{R^{q – 1}},\;c = A{R^{r – 1}}$

$\therefore $${\left( {\frac{c}{b}} \right)^p}{\left( {\frac{b}{a}} \right)^r}{\left( {\frac{a}{c}} \right)^q} = {\left( {\frac{{A{R^{r – 1}}}}{{A{R^{q – 1}}}}} \right)^p}{\left( {\frac{{A{R^{q – 1}}}}{{A{R^{p – 1}}}}} \right)^r}{\left( {\frac{{A{R^{p – 1}}}}{{A{R^{r – 1}}}}} \right)^q}$

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

Standard 11

Mathematics

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

Solution

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