Today the integer ages of brothers Alex, Bert, and Carl form an arithmetic progression.
Two months later, after Alex's birthday, the ages form a geometric progression.
Is Alex the oldest, youngest, or middle child?
You can't tell.
Quick proof: suppose the three ages after Alex's birthday are a^2, a(a+1), and (a+1)^2. Then they're a geometric progression with ratio (a+1)/a. The difference between the first two is a and the difference between the 2nd two is (a+1).
If Alex were the youngest, then two months ago Alex's age would be a^2-1 and the ages would be an arithmetic progression with difference (a+1). But if Alex were the oldest, then two months ago Alex would have been a^2 + 2a and the ages would be an arithmetic progression with difference a.
For example, if Alex is youngest, then 8, 12, 16 => 9, 12, 16, but if Alex is oldest, then 9, 12, 15 => 9, 12, 16
You *can* say that Alex isn't the middle child, though.
Suppose the three ages before Alex's birthday were a, a+r and a+2r. If Alex is the middle child, then after two months the ages are a, a+r+1 and a+2r. If this is a geometric progression, then
a(a+2r) = (a+r+1)^2
a^2 + 2ar = a^2 + 2ar + a + r^2 + 2r + 1
0 = a + r^2 + 2r + 1
-a = (r+1)^2
Since the left side is < 0 and the right side is > 0, there are no solutions.
Posted by Paul
on 2017-01-16 12:39:18