Q.

Let a1,a2,a3,............. be a sequence of positive integers in arithmetic progression with common difference 2. Also, let b1,b2,b3......... be a sequence of positive integers in geometric progression with common ratio 2. If a1=b1=c, then the number of all possible values of c, for which the equality 2(a1+a2+......+an)=b1+b2+........+bn holds for some positive integer n, is ___________

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answer is 1.

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Detailed Solution

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2(a1+a2+........+an)=b1+b2+..........+bn2(n2(2c+(n1)2))=c(2n121)

c=2n(n1)2n12n

2x2>x2x if  x8

2x1>2x if  x>3

So, c<2n12n1n8

as c1 values of c are not possible for n8

possible values of n = 3, 4, 5, 6, 7; only satisfy when n = 3 and c = 12

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