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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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The product of 3 numbers in a GP is 64 and the sum of product of numbers taken in pairs is 56. Find the three numbers in GP.

$\begin{array}{1 1} 2,4,8 \\ 8,4,2 \\ both (A) \;and \;(B) \\ 1,4,16 \end{array}$

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Answer : (c) $ both\;a\;\xi\;b$
Explanation : Let the 3 numbers in GP be a/r, a, ar
$a/r*a*ar=64\;\qquad a^3=64$
a=4
$Given\;a/r*a+a*ar+a/r*ar=56$
$a^2/r+a^2\;r+a^2=56\;\qquad\;a=4\;\quad a^2=16$
$16/r+16r+16=56\;\qquad\;2r^2-5r+2=0$
$2r^2-r-4r+2=0\;\qquad\;2r(r-2)-1(r-2)=0$
$2r-1=0\;\quad\;or\;\quad\;r-2=0$
$r=1/2\;\quad\;or\;\quad\;r=2$
$a/r,a,ar=8,4,2\;\quad\;or\;\quad\;2,4,8.$
answered Dec 31, 2013 by yamini.v
 

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