2022 AMC 10A Problem 20

Let the arithmetic sequence be $a, a+d, a+2d, a+3d$ and the geometric sequence $b, br, br^2, br^3$, all positive integers. Then
$$a + b = 57, \qquad a + d + br = 60, \qquad a + 2d + br^2 = 91.$$
Subtract consecutive equations to eliminate $a$:
$$d + b(r-1) = 3, \qquad d + br(r-1) = 31.$$
Subtract again to eliminate $d$: $b(r-1)^2 = 28$.

The ratio $r$ is a positive integer greater than $1$ (a fractional ratio $p/q$ would need $q^3 \mid b$ with $b < 57$, and $q = 2, 3$ fail quickly). So $(r-1)^2$ is a square divisor of $28$: $1$ or $4$.

• $r = 2$: $b = 28$, $d = 3 - 28 = -25$, $a = 29$. The arithmetic terms $29, 4, -21, \dots$ are not all positive. Rejected.
• $r = 3$: $b = 7$, $d = 3 - 14 = -11$, $a = 50$. Arithmetic: $50, 39, 28, 17$; geometric: $7, 21, 63, 189$. Check: $57, 60, 91$. Valid.

The fourth term is $17 + 189 = 206$.

The answer is $\boxed{\textbf{(E)}\ 206}$.

Differences of consecutive terms kill the constant of an arithmetic sequence, and second differences kill the common difference too, isolating the geometric part as $b(r-1)^2$. The integer condition then leaves only a couple of candidate ratios, and the positivity condition selects one. Always verify the reconstructed sequences against the given data.

• Accepting r = 2 (b = 28, d = -25), which makes the arithmetic sequence 29, 4, -21, ... negative; all terms must be positive integers, forcing r = 3.
• Assuming the arithmetic sequence must be increasing; here $d = -11$ and the two sequences move in opposite directions.