JEE PYQ: Sequence And Series Question 12
Question 12 - 2021 (25 Feb Shift 1)
Let $A_1, A_2, A_3, \ldots$ 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
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Answer: (9)
Solution
$x = \frac{12}{\sqrt{2}}$, $y = \frac{12}{(\sqrt{2})^2}$
$\therefore$ Side lengths are in G.P.
$T_n = \frac{12}{(\sqrt{2})^{n-1}}$
$\therefore$ Area $= \frac{144}{2^{n-1}} < 1 \Rightarrow 2^{n-1} > 144$
Smallest $n = 9$
BETA
