1982 AHSME Problems/Problem 8
Problem
By definition, and , where are positive integers and . If form an arithmetic progression with , then equals
Solution
We know that because they form an arithmetic sequence, and expanding, we have by the definitions in the problem:
Canceling out the and the from each side of the equals sign, we have Getting rid of the fractions by cross multiplication, and getting n on one side, we have and we can factor out the n, so n(n^2-9n+14)=0, and we are looking for two integers x and y such that and By guess and check, our integers are -7 and -2, so According to the problem, so we have n=7 or 2, thus
~ab2024