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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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The sum of three terms of a strictly increasing GP is $\alpha S$ and sum of their squares is $S^{2}$. $\alpha^{2}$ lies in

$(a)\;(\frac{1}{3},2)\qquad(b)\;(\frac{1}{3},3)\qquad(c)\;(1,2)\qquad(d)\;None\;of\;these$

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Answer : (d) None of these
Explanation : $a\;(\frac{1}{r}+1+r)=\alpha\;S-----(1)$
$a^2\;(\frac{1}{r^2}+1+r^2)=S^{2}------(2)$
Dividing (2) by (1)
$a\;(\frac{1}{r}-1+r)=\large\frac{S}{\alpha}-------(3)$
From (2) and (3)
$2a=S\;(\alpha-\large\frac{1}{\alpha})=S\;(\large\frac{\alpha^2-1}{\alpha})$
Putting this in (2) we get ,
$\large\frac{(\alpha^2-1)^2}{4\alpha^2}\;(\large\frac{1}{r^2}+1+r^2)=1$
$(r-\large\frac{1}{r})^2+3=\large\frac{4\alpha^2}{(\alpha^2-1)^2}$
$3\alpha^4-10\alpha^2+3 < 0$
$(3\alpha^2-1)(\alpha^2-3) < 0$
$\large\frac{1}{3} < \alpha^2 < 3$
But $\alpha^2=1\quad\;a=0\;is\;not\;possible$
$so\;,\alpha^2 \in (\large\frac{1}{3},3)----(1)\;.$
answered Jan 23, 2014 by yamini.v
 

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