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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If 20 is divided into four parts which are in AP such that ratio of product of first and fourth part and product of second and third part is 2:3, then the largest part is

$(a)\;6\qquad(b)\;7\qquad(c)\;8\qquad(d)\;9$

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Answer : (c) 8
Explanation : Let the four terms be $\;a-3d\;,a-d\;,a+d\;,and\;a+3d$ with common difference 2d.
$sum=4a=20$
$a=5$
$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{2}{3}$
$\frac{25-9d^2}{25-d^2}=\frac{2}{3}$
$d^2=1$
$d=1$
$Largest \;term=a+3d$
$=5+3(1)$
$=8.$
answered Jan 21, 2014 by yamini.v
 

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