In the 2022 AMC 10A competition, Problem #20 combines two fundamental sequences—arithmetic and geometric—into a multi-constraint system:

Contest Problem Statement:
"A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are 57, 60, and 91. What is the fourth term of this sequence?"

Setting Up the Equations

Let the terms of the two sequences be:

• Arithmetic progression (AP): $A_1 = a, \; A_2 = a + d, \; A_3 = a + 2d, \; A_4 = a + 3d$.
• Geometric progression (GP): $G_1 = g, \; G_2 = gr, \; G_3 = gr^2, \; G_4 = gr^3$.

Every term in both sequences must be a positive integer ($a, a+d, a+2d, a+3d \ge 1$ and $g, gr, gr^2, gr^3 \ge 1$).

The problem gives the first three combined terms:

1. $A_1 + G_1 = a + g = 57 \implies a = 57 - g$.
2. $A_2 + G_2 = a + d + gr = 60 \implies d + g(r - 1) = 3$.
3. $A_3 + G_3 = a + 2d + gr^2 = 91 \implies 2d + g(r^2 - 1) = 34$.

The Algebraic Breakthrough

From equation (2), we have $2d = 6 - 2g(r - 1)$. Substituting this into equation (3): $$6 - 2g(r - 1) + g(r^2 - 1) = 34$$ $$g(r^2 - 2r + 1) = 28$$ Notice that $r^2 - 2r + 1 = (r - 1)^2$, so: $$g(r - 1)^2 = 28$$ Since $g$ is an integer and $(r - 1)^2$ must be a perfect square dividing $28 = 4 \times 7$:

• If $(r - 1)^2 = 1 \implies r = 2, g = 28$: Then $a = 57 - 28 = 29$, and $d = 3 - 28(1) = -25$. The third arithmetic term would be $29 + 2(-25) = -21$, which is negative (violating the positive integer rule)!
• If $(r - 1)^2 = 4 \implies r - 1 = 2 \implies r = 3, g = 7$: Then $a = 57 - 7 = 50$, and $d = 3 - 7(2) = -11$. The arithmetic terms are $50, 39, 28, 17$ — all positive! The geometric terms are $7, 21, 63, 189$ — all positive!

Summing the fourth terms: $17 + 189 = 206$.

Solving with Prolog

In Prolog, we express the domain boundaries and arithmetic relationships directly. Prolog evaluates the system, filters out invalid negative terms, and outputs the fourth term.

Prolog prunes candidate branches where terms turn negative, cleanly pinpointing the unique valid progression.