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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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In a GP, the ratio of the sum of first 3 terms to the sum of first 6 terms is 125/152. Find the common of the GP.

$1 \\ 5/6 \\ 3/5 \\ 6/5 $

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Answer : (c) 3/5
$Explanation : S_{n}=a\;\frac{r^n-1}{r-1}\;in\;GP$
$Given \;that $
$\frac{S_{3}}{S_{6}}=\frac{a(r^3-1/r-1)}{a(r^6-1/r-1)}=125/152$
$Assuming \;r\;\neq\;1$
$(as\; r-1\;is\;in\;denominator\;\xi\;denomitor\;cannot\;be\;0\;,r-1\;\neq\;0,\;r\;\neq\;1)$
$\frac{a(r^3-1/r-1)}{a(r^6-1/r-1)}=125/152$
$\frac{r^3-1}{r^6-1}=125/152$
$152r^3-152=125r^6-125$
$125r^6-152r^3-27=0$
$125r^3(r^3-1)-27(r^3-1=0)$
$(125r^3-27)(r^3-1)=0$
$125r^3=27\quad\;or\;\quad\;r^3-1=0$
$r^3=27/125\;\quad\;r=1$
$r=3/5\;\quad\;r\neq\;1$
$Answer\;=\;r=3/5\;is\;common\;ratio.$
answered Dec 31, 2013 by yamini.v
 

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