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
Answer: 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, a2 = 1 + nd, then (1 + md)2 = 1 + nd, so d = (n - 2m)/m2, 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 cn, 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.
© John Scholes
jscholes@kalva.demon.co.uk
9 Sep 2002