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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If a, b, c are in GP, find the value of $a^2b^2c^2(\large\frac{1}{a^3}+\large\frac{1}{b^3}+\large\frac{1}{c^3})$

$\begin{array}{1 1} a^3 b^3c^3 \\ 1\\\frac{a^3+b^3+c^3}{a^2b^2c^2} \\a^3+b^3+c^3\end{array}$

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1 Answer

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Answer : $(d)\;a^3+b^3+c^3$
Explanation : a,b,c are in GP $ \;b^2=ac$
$a^2b^2c^2\;(\large\frac{1}{a^3}+\large\frac{1}{b^3}+\large\frac{1}{c^3})=a^2(ac)c^2(\large\frac{1}{a^3}+\large\frac{1}{b^3}+\large\frac{1}{c^3})$
$=a^3c^3(\large\frac{1}{a^3}+\large\frac{1}{b^3}+\large\frac{1}{c^3})$
$=c^3+\large\frac{a^3c^3}{b^3}+a^3\qquad\;b^2=ac$
$=c^3+\large\frac{b^6}{b^3}+a^3$
$=c^3+b^3+a^3$
$=a^3+b^3+c^3.$
answered Jan 3, 2014 by yamini.v
edited Jan 24, 2014
 

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