Paragraph: Let $M=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^{2}+y^{2} \leq r^{2}\right\}$, where…

Paragraph: Let $M=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^{2}+y^{2} \leq r^{2}\right\}$, where $r>0 .$ Consider the geometric progression $a_{n}=\frac{1}{2^{n-1}}, n=1,2,3, \ldots .$ 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 $\left(S_{n-1}, 0\right)$ and radius $a_{n}$, and $D_{n}$ denote the circle with center $\left(S_{n-1}, S_{n-1}\right)$ and radius $a_{n}$.
Question: 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
  1. k+2l=22
  2. 2k+l=26
  3. 2k+3l=34
  4. 3k+2l=40

Solution

Sn=1+12+122++12n-1

=21-12n=2-12n-1

Centre of Cn is 2-12n-2,0

and radius of Cn is 12n-1

when r=1025 513<2

Cn will lie inside m

when 2-12n-2+12n-1<1025 513

1-12n<10251026

2n<1026n10

Hence number of circles

k=10

Also =5

3k+2=30+10=40

Asked in: JEE Advanced 2021 (Paper 2)

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