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Sequence Group (Posted on 2003-09-23) Difficulty: 3 of 5
I didn't come up with this problem, but I still think it's a good one.

There are 4 positive integers in order from least to greatest, such that the first three make an aritmetic sequence, and the last three make a geometric sequence. If the difference between the largest and smallest term is 30, what are the terms?

See The Solution Submitted by Gamer    
Rating: 3.4000 (5 votes)

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Solution Puzzle Solution Comment 8 of 8 |
(In reply to Answer by K Sengupta)

Since the four terms are distinct, and the first three terms are in
arithmetic sequence, we must have the first three terms as y-b, y and y+b. Since all the four terms are positive, it follows that y> b.

Now, the last three terms are in geometric sequence. Thus, the common ratio is (y+b)/y, so that the 4th term is (y+b)^2/y. Now, the  difference between the last term and the first is 30, and so:

(y+b)^2/y - (y-b) = 30
or, b(3 + b/y) = 30 ......(*)

Since b< y, it follows that:

30 = b(3 + b/y) < 4*b, and:

30 = b(3 + b/y) > 3*b, so that:

7.5< b< 10.

The only values that satisfy the above inequality occur at b = 8, 9

Substituting b = 8 in (*), we have:

24 + 64/y = 30, or y = 32/3, which is not an integer and thus leads to a contradiction.

Substituting b = 9 in (*), we have:

24 + 81/y = 30, or y = 27, so that;
(y+b)^2/y = 36^2/27 = 48, and (y-b, y, y+b) = (18, 27, 36)

Thus, the required four terms of the given sequence are
18, 27, 36 and 48.

  Posted by K Sengupta on 2008-06-07 06:01:00
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