| Arithmetic and Geometric |
| Four distinct positive integers (A1, A2, A3, and A4) form an arithmetic sequence. A1, A2, and A4 form a geometric sequence. The sum of the four terms is a perfect cube. What is the smallest possible value of A1? |
Problem Moderated by: Douglas |
| Problem Solution |
Call the terms a-2d, a-d, a, and a+d.
Now, (a+d)/(a-d)=(a-d)/(a-2d), or a=3d. The terms of the sequence, then, are d, 2d, 3d, and 4d, which sums to (2)(5)d, which must be a perfect cube - therefore, the value of the first term, d must be (2^2)(5^2)=100.
Solution submitted by Sasha J. |
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