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Question

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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Solution

2(a1+a2++an)=b1+b2++bn
2[n2(2a1+(n1)2)]=b1(2n1)21
2n[a1+(n1)]=b1(2n1)
2na1+2n22n=a1(2n1)(a1=b1)
2n22n=a1(2n12n)
a1=2(n2n)(2n12n)=c (a1=c)
c1
therefore, n=1,2,3,4,5,6
n=1c=0(rejected)
n=2c<0(rejected)
n=3c=12(correct)
n=4c=not Integer
n=5c=not Integer
c=12 for n=3
Hence, no. of such c=1

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