 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Age Progression (Posted on 2017-01-16) 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?

 No Solution Yet Submitted by Brian Smith No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution? | Comment 1 of 3
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 Please log in:

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