Problem:
A sequence of positive integers with a1=1 and a9+a10=646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all n≥1, the terms a2n−1,a2n, and a2n+1 are in geometric progression, and the terms a2n, a2n+1, and a2n+2 are in arithmetic progression. Let an be the greatest term in this sequence that is less than 1000 . Find n+an.
Solution:
The terms in the sequence are 1,r,r2,r(2r−1),(2r−1)2,(2r−1)(3r−2),(3r− 2)2,…. Assuming that the pattern continues, the ninth term is (4r−3)2 and the tenth term is (4r−3)(5r−4). Thus (4r−3)2+(4r−3)(5r−4)=646. This leads to (36r+125)(r−5)=0. Because the terms are positive, r=5. Substitute to find that an=(2n−1)2 when n is odd, and that an=(2n−3)(2n+1) when n is even. The least odd-numbered term greater than 1000 is therefore a17=332=1089, and a16=29⋅33=957<1000. The desired value of n+an is 957+16=973.
The pattern referred to above is
a2na2n+1=[(n−1)r−(n−2)][nr−(n−1)]=[nr−(n−1)]2
This pattern has been verified for the first few positive integral values of n. The above equations imply that
a2n+2a2n+3=2[nr−(n−1)]2−[(n−1)r−(n−2)][nr−(n−1)]=[nr−(n−1)][2nr−2(n−1)−(n−1)r+(n−2)]=[nr−(n−1)][(n+1)r−n], and =[nr−(n−1)]2[nr−(n−1)]2[(n+1)r−n]2=[(n+1)r−n]2
The above argument, along with the fact that the pattern holds for n=1 and n=2, implies that it holds for all positive integers n.
The problems on this page are the property of the MAA's American Mathematics Competitions