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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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Let $S_{1}, S_{2}$....be the squares such that for each $n\geq\;1$, the length of side of $S_{n}$ = length of diagonal of $S_{n+1}$. If side of $S_{1}$ is 20 cm, the smallest value of n such that area of $S_{n}\;<\;2$ is

$(a)\;7\qquad(b)\;8\qquad(c)\;9\qquad(d)\;10$

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Answer : (c) 9
Explanation : Let $\;a_{n}\;be\;the\;length\;of\;side$
$a_{n}=\sqrt{2}\;a_{n+1}$
$\frac{a_{n+1}}{a_{n}}=\frac{1}{\sqrt{2}}--------- \;GP$
$So\;,\;a_{n}=20\;(\frac{1}{\sqrt{2}})^{n-1}$
$Area\;=a_{n}^2=400\;\frac{1}{2^n-1}\;<\;2$
$400\;<\;2^n$
$n=9\;$ (smallest possible n.
answered Jan 20, 2014 by yamini.v
 

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