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Question

Let M={(x,y)R×R:x2+y2r2}, where r>0. Consider the geometric progression an=12n1,n=1,2,3, . Let S0=0 and, for n1, let Sn denote the sum of the first n terms of this progression. For n1 let Cn denote the circle with center (Sn1,0) and radius an and Dn denote the circle with center (Sn1,Sn1) and radius an.

Consider M with r=1025513. Let k be the number of all those circles Cn that are inside M. Let l be the maximum possible number of circles among these k circles such that no two circles intersect. Then

A
k+2l=22
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B
2k+l=26
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C
2k+3l=34
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D
3k+2l=40
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Solution

The correct option is D 3k+2l=40
an=12n1 and Sn=2(112n)
For circles Cn to be inside M.
Sn1+an<1025513
Sn<1025513
112n<10251026=111026
2n<1026
n10
Number of circles inside M be 10=k
Clearly alternate circles do not intersect each other i.e., C1,C3,C5,C7,C9 do not intersect each other as well as C2,C4,C6,C8 and C10 do not intersect each other hence maximum 5 set of circles do not intersect each other.
l=5
3k+2l=40

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