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+(n−1)2)]=b1(2n−1)2−1 ⇒2n[a1+(n−1)]=b1(2n−1) ⇒2na1+2n2−2n=a1(2n−1)(∵a1=b1) ⇒2n2−2n=a1(2n−1−2n) a1=2(n2−n)(2n−1−2n)=c (∵a1=c) ∵c≥1 therefore, n=1,2,3,4,5,6 n=1⇒c=0(rejected) n=2⇒c<0(rejected) n=3⇒c=12(correct) n=4⇒c=not Integer n=5⇒c=not Integer ∴c=12 for n=3 Hence, no. of such c=1

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