Problem:
A strictly increasing sequence of positive integers a1,a2,a3,… has the property that for every positive integer k, the subsequence a2k−1,a2k,a2k+1 is geometric and the subsequence a2k,a2k+1,a2k+2 is arithmetic. Suppose that a13=2016. Find a1.
Solution:
The ratio a1a2 must be rational, so let a2=aba1, where a and b are relatively prime positive integers and a<b. Because a3=a2b2a1 is also an integer, there is an integer c such that a1=ca2 and a2=cab. Thus the sequence begins ca2, cab,cb2,cb(2b−a). Examining a few terms of the sequence suggests that for k≥1
a2k−1=c((k−1)b−(k−2)a)2 and a2k=c((k−1)b−(k−2)a)(kb−(k−1)a).
Then, because the sequence a2k−1,a2k,a2k+1 is geometric, it would follow that
a2k+1=a2k−1a2k2=c((k−1)b−(k−2)a)2c2((k−1)b−(k−2)a)2(kb−(k−1)a)2=c(kb−(k−1)a)2.
It would then follow that for k≥1 that
a2k=c((k−1)b−(k−2)a)(kb−(k−1)a) and a2k+1=c(kb−(k−1)a)2.
Because a2k,a2k+1,a2k+2 is arithmetic, it would follow that
a2k+2=2a2k+1−a2k=2c(kb−(k−1)a)2−c((k−1)b−(k−2)a)(kb−(k−1)a)=c(kb−(k−1)a)((k+1)b−ka).
It can now be verified by mathematical induction that for all positive integers k
a2k=c((k−1)b−(k−2)a)(kb−(k−1)a) and a2k+1=c(kb−(k−1)a)2.
In particular, a13=c(6b−5a)2=2016=14⋅122. Therefore 6b−5a is a factor of 12 and is also the seventh term in an arithmetic progression whose first two terms are a and b. Let n=6b−5a. Then a<a+6(b−a)=n, and 6b=5a+n≡n−a(mod6), implying that n−a is a multiple of 6. Thus 6<a+6≤n≤12, and the only solution for (a,b,n) in positive integers is (6,7,12). The corresponding value of c is 14, and a1=14⋅62=504.
The problems on this page are the property of the MAA's American Mathematics Competitions