Consider with . Let be the number of all those circles that are inside . Let be the maximum possible number of circles among these circles such that no two circles intersect. Then [2021]

Question

Let $M = {(x, y) \in ℝ \times ℝ : x^{2} + y^{2} \leq r^{2}}$ where $r > 0$ .

Consider the geometric progression $a_{n} = \frac{1}{2^{n - 1}}, n = 1, 2, 3, \dots$ . Let $S_{0} = 0$ and, for $n \geq 1$ , let $S_{n}$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$ , let $C_{n}$ denote the circle with center $(S_{n - 1}, 0)$ and radius $a_{n}$ and $D_{n}$ denote the circle with center $(S_{n - 1}, S_{n - 1})$ and radius $a_{n}$ . [2021]

Q . Consider $M$ with $r = \frac{1025}{513}$ . Let $k$ be the number of all those circles $C_{n}$ that are inside $M$ . Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then

Smart Achievers · Accepted Answer

(4)

$∵$ $a_{n} = \frac{1}{2^{n - 1}}$

$S_{n} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n - 1}} = 2 (1 - \frac{1}{2^{n}}) = 2 - \frac{1}{2^{n - 1}}$

$S_{n}$

$S_{n - 1} + a_{n} < \frac{1025}{513} \Rightarrow 2 - \frac{1}{2^{n - 2}} + \frac{1}{2^{n - 1}} < \frac{1025}{513}$

$\Rightarrow 1 - \frac{1}{2^{n}} < \frac{1025}{1026} = 1 - \frac{1}{1026}$

$\Rightarrow 2^{n} < 1026 \Rightarrow n \leq 10 \Rightarrow k = 10$

$Also l = 5$

$3 k + 2 l = 30 + 10 = 40$

Solution

1 $k + 2 l = 22$

2 $2 k + l = 26$

3 $2 k + 3 l = 34$

4 $3 k + 2 l = 40$

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Solution

1 k+2l=22

2 2k+l=26

3 2k+3l=34

4 3k+2l=40

Ans. (4) ∵ an=12 n-1 Sn=1+12+122+⋯+12n-1=2(1-12n)=2-12 n-1 For circles Cn to lie inside M Sn-1+an<1025513⇒2-12 n-2+12 n-1<1025513 ⇒1-12n<10251026=1-11026 ⇒2n<1026⇒n≤10⇒k=10 Also l=5 3k+2l=30+10=40

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1 $k + 2 l = 22$

2 $2 k + l = 26$

3 $2 k + 3 l = 34$

4 $3 k + 2 l = 40$