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?

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Arithmetic and Geometric 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 (2²)(5²)=100.

Solution submitted by Sasha J.

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