2022 AMC 10A Problem 20

Step 1: name the sequences. Let the arithmetic sequence be $a,\ a + d,\ a + 2d,\ a + 3d$ and the geometric one $b,\ br,\ br^2,\ br^3$. Adding term by term:
$$a + b = 57, \qquad a + d + br = 60, \qquad a + 2d + br^2 = 91.$$

Step 2: subtract neighbors to remove $a$.
- Second minus first: $d + br - b = 3$.
- Third minus second: $d + br^2 - br = 31$.

Step 3: subtract again to remove $d$. $br^2 - 2br + b = 28$, which factors as
$$b(r - 1)^2 = 28.$$

Step 4: which ratios are possible? First, $r$ is a whole number. (If $r$ were a fraction with bottom $q$ in lowest terms, then $br^3$ being whole would need $b$ to be a multiple of $q^3$. Since $b < 57$, $q$ could only be $2$ or $3$, and checking those few values of $b$ shows none of them gives $b(r-1)^2 = 28$.) Also $r = 1$ would give $0 = 28$. So $r \ge 2$, and $(r - 1)^2$ is a perfect square that divides $28$: it is $1$ or $4$. So $r = 2$ or $r = 3$.

Step 5: test both.
- $r = 2$: $b = 28$, then $d = 3 - 28 = -25$ and $a = 57 - 28 = 29$. The arithmetic terms are $29, 4, -21, \ldots$, and $-21$ is not positive. Rejected.
- $r = 3$: $b = 28 \div 4 = 7$, then $d = 3 - (21 - 7) = -11$ and $a = 50$. Arithmetic: $50, 39, 28, 17$. Geometric: $7, 21, 63, 189$. Sums: $57, 60, 91$, as required.

Step 6: the fourth term. $17 + 189 = 206$.

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

Taking differences of neighboring terms removes the starting value of an arithmetic sequence, and taking differences again removes its common difference too. What is left belongs only to the geometric part, and whole-number conditions leave just a couple of ratios to test. Always rebuild both sequences at the end and check every condition, including "positive".

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 $\boxed{\textbf{(E)}\ 206}$.