15th Iberoamerican 2000 Problem B1 Some terms are deleted from an infinite arithmetic progression 1, x, y, ... of real numbers to leave an infinite geometric progression 1, a, b, ... . Find all possible values of a. Solution
: the positive integers. If a is negative, then the terms in the GP are alternately positive and negative, whereas either all terms in the AP from a certain point on are positive or all terms from a certain point on are negative. So a cannot be negative. If a is zero, then all terms in the GP except the first are zero, but at most one term of the AP is zero, so a cannot be zero. Thus a must be positive, so the AP must have infinitely many positive terms and hence x ≥ 1. Let d = x - 1, so all terms of the AP have the form 1+nd for some positive integer n. Suppose a = 1 + md, a 2 = 1 + nd, then (1 + md) 2 = 1 + nd, so d = (n - 2m)/m 2 , which is rational. Hence a is rational. Suppose a = b/c, where b and c are relatively prime positive integers and c > 1. Then the denominator of the nth term of the GP is c n , which becomes arbitrarily large as n increases. But if d = h/k, then all terms of the AP have denominator at most k. So we cannot have c > 1. So a must be a positive integer. On the other hand, it is easy to see that any positive integer works. Take x = 2, then the AP includes all positive integers and hence includes any GP with positive integer terms. 15th Ibero 2000 © John Scholes jscholes@kalva.demon.co.uk 9 Sep 2002