Let A_{1}, A_{2}, A_{3}, ldots be squares suc | vedclass

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(D) Let $a_{n}$ be the side length of square $A_{n}$.Given that the side length of $A_{n}$ equals the diagonal of $A_{n+1}$,we have $a_{n} = \sqrt{2}

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let $a_{n}$ be the side length of square $A_{n}$.Given that the side length of $A_{n}$ equals the diagonal of $A_{n+1}$,we have $a_{n} = \sqrt{2} a_{n+1}$,which implies $a_{n+1} = \frac{a_{n}}{\sqrt{2}}$.This forms a geometric progression for the side lengths with first term $a_{1} = 12$ and common ratio $r = \frac{1}{\sqrt{2}}$.The side length $a_{n}$ is given by $a_{n} = a_{1} 	imes r^{n-1} = 12 	imes \left(\frac{1}{\sqrt{2}}ight)^{n-1}$.The area of $A_{n}$ is $(a_{n})^2 = 144 	imes \left(\frac{1}{2}ight)^{n-1} = \frac{144}{2^{n-1}}$.We want the area to be less than $1$,so $\frac{144}{2^{n-1}} This implies $2^{n-1} > 144$.Since $2^{7} = 128$ and $2^{8} = 256$,we must have $n-1 \geq 8$.Therefore,$n \geq 9$. The smallest value of $n$ is $9$.

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