China Mathematical Competition

Question

Let a_n be an arithmetic progression with common difference d (d 0) and b_n be a geometric progression with common ratio q, where q is a positive rational number less than 1. If a_1 = d, b_1 = d^2 and a_1^2 + a_2^2 + a_3^2b_1 + b_2 + b_3 is a positive integer, then q equals ______.

Accepted Answer

As aligned a_1^2 + a_2^2 + a_3^2b_1 + b_2 + b_3 &= a_1^2 + (a_1+d)^2 + (a_1+2d)^2b_1 + b_1q + b_1q^2 &= 141+q+q^2 = m aligned is a positive integer, we get 1 + q + q^2 = 14m. Then q = -12 + 14 + 14m - 1 = -12 + 56-3m4m Since q is a positive rational number less than 1, we have 1 < 14m < 3, i.e. 5 m 13, and 56-3m4m is the square of a rational number. We can verify that only m = 8 meets the required. That means q = 12.