 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Four From Arithmetic and Geometric (Posted on 2016-04-29) Each of X, Y and Z is a distinct positive integer such that
X, Y and Z are in arithmetic sequence, and
X, Y and Z+2016 are in geometric sequence.

Find the four smallest values of X+Y+Z.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Infinitely more solutions | Comment 2 of 3 | Actually, there are an infinite number of solutions in positive integers.

Starting with Jer's y=x(+/-)x*sqrt(2016/x)

The (-) gives positive solutions if x = 2016*n^2
Then y = 2016(n^2 - n)
z = 2016(n^2 - 2n)

Checking, we see that 2016*(n^2, n*(n-1), (n-1)^2) is indeed a geometric progression, with ratio (n-1)/n

In order for z to be positive, n must be >= 3

The first of these solutions is 2016*(9,6,3).
The next is 2016*(16,12,8).
etc.

Obviously, none of these are among the 4 smallest

 Posted by Steve Herman on 2016-04-30 08:35:14 Please log in:

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