Sequence And Series Question 60
Question: Let $ S_1,\ S_2,……. $ be squares such that for each $ n\ge 1 $ , the length of a side of $ S_{n} $ equals the length of a diagonal of $ {S_{n+1}} $ . If the length of a side of $ S_1 $ is $ 10cm $ , then for which of the following values of $ n $ is the area of $ S_{n} $ less then $ 1\ sq\ cm $
[IIT 1999]
Options:
A) 7
B) 8
C) 9
D) 10
Show Answer
Answer:
Correct Answer: C
Solution:
Given $ x_{n}={x_{n+1}}\sqrt{2} $
$ \therefore $ $ x_1=x_2\sqrt{2},\ x_2=x_3\sqrt{2},\ x_{n}={x_{n+1}}\sqrt{2} $ On multiplying $ x_1={x_{n+1}}{{(\sqrt{2})}^{n}} $
$ \Rightarrow $ $ {x_{n+1}}=\frac{x_1}{{{(\sqrt{2})}^{n}}} $ Hence $ x_{n}=\frac{x_1}{{{(\sqrt{2})}^{n-1}}} $ Area of $ S_{n}=x_n^{2}=\frac{x_1^{2}}{{2^{n-1}}}<1 $
$ \Rightarrow $ $ {2^{n-1}}>x_1^{2}\ \ \ \ \ \ (x_1=10) $
$ \therefore $ $ {2^{n-1}}>100 $ But $ 2^{7}>100,\ 2^{8}>100,\ $ etc.
$ \therefore $ $ n-1=7,\ 8,\ 9….. $
$ \Rightarrow $ $ n=8,\ 9,\ 10….. $
BETA
