JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(A_{1}, A_{2}, A_{3}, \ldots \ldots . .\) be squares such that for each \(n \geq 1,\) the length of the side of \(A _{ n }\) equals the length of diagonal of \(A _{ n +1}\). If the length of \(A _{1}\) is \(12\, cm ,\) then the smallest value of \(n\) for which area of \(A _{ n }\) is less than one, is ..........
- A \(8\)
- B \(6\)
- C \(3\)
- D \(9\)
Answer & Solution
Correct Answer
(D) \(9\)
Step-by-step Solution
Detailed explanation
Let \(a_{n}\) be the side length of \(A_{n}\). So, \(a _{ n }=\sqrt{2} a _{ n +1}, a _{1}=12\) \(\Rightarrow a _{ n }=12 \times\left(\frac{1}{\sqrt{2}}\right)^{ n -1}\) Now, \(\left(a_{n}\right)^{2}<1 \Rightarrow \frac{144}{2^{(n-1)}}<1\) \(\Rightarrow 2^{(n-1)}>144\)…
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