2025 AMC 10A Problem 11

The two sequences share only the number $z$. So find out what each sequence says about $z$, then find the smallest $z$ that satisfies both.

1. The geometric sequence $1, p, q, z$. Each term is the one before times the same ratio. Starting from $1$, the ratio is $p$ itself, so the terms are
$$1,\ p,\ p^2,\ p^3.$$
Since $p$ is a term, it is a whole number, and "increasing" means $p$ is at least $2$. So $z = p^3$ is a perfect cube: $8, 27, 64, \ldots$

2. The arithmetic sequence $1, x, y, z$. Each step adds the same whole number $d$. So the terms are $1,\ 1 + d,\ 1 + 2d,\ 1 + 3d$, and
$$z = 1 + 3d.$$
That means $z - 1$ must be a multiple of $3$.

1. Test cubes from smallest up.

So the smallest possible $z$ is $64$.

1. Write out both sequences. With $d = 21$ the arithmetic one is $1, 22, 43, 64$. With $p = 4$ the geometric one is $1, 4, 16, 64$. Both are increasing and made of whole numbers.

5. Add up what is asked.
$$x + y + z + p + q = 22 + 43 + 64 + 4 + 16 = 149.$$

The answer is $\boxed{\textbf{(E)}\ 149}$.

A geometric sequence that starts at $1$ shows its ratio as the second term, so "all whole numbers" forces a whole-number ratio and makes the last term a cube. When two sequences share one term, turn each sequence into a rule about that term, then search upward for the smallest number that obeys both rules. That is the usual way to handle "as small as possible" questions.

Find a formula for the total first. In the arithmetic sequence, $x + y = (1 + d) + (1 + 2d) = 2 + 3d$, which is $1 + z$. In the geometric one, $p + q = p + p^2$. So
$$x + y + z + p + q = 1 + 2z + p + p^2.$$
For $p = 2$ this is $1 + 16 + 2 + 4 = 23$, and for $p = 3$ it is $1 + 54 + 3 + 9 = 67$. Neither is a choice, which agrees with those cubes failing. For $p = 4$ it is $1 + 128 + 4 + 16 = 149$, choice (E).

The geometric sequence. Its first term is $1$, so its ratio is $p/1 = p$, an integer, and the sequence is $1, p, p^2, p^3$ with $z = p^3$. Strictly increasing forces $p \ge 2$.

The arithmetic sequence. With common difference $d$ the terms are $1, 1+d, 1+2d, 1+3d$, so
$$z = 1 + 3d, \qquad\text{that is}\qquad z \equiv 1 \pmod 3,$$
and $d \ge 1$.

So $z$ must be a cube $p^3$ with $p \ge 2$ and $p^3 \equiv 1 \pmod 3$. Since $p^3 \equiv p \pmod 3$ (check $p \equiv 0, 1, 2$ in turn), this says simply $p \equiv 1 \pmod 3$. Test in order:

• $p = 2$: $z = 8$, and $8 - 1 = 7$ is not a multiple of $3$. Fails.
• $p = 3$: $z = 27$, and $27 - 1 = 26$ is not a multiple of $3$. Fails.
• $p = 4$: $z = 64$, and $64 - 1 = 63 = 3 \cdot 21$. Works, with $d = 21$.

The smallest possible $z$ is therefore $64$, giving $1, 22, 43, 64$ and $1, 4, 16, 64$. Both are strictly increasing sequences of integers, so
$$x + y + z + p + q = 22 + 43 + 64 + 4 + 16 = 149.$$

The answer is $\boxed{\textbf{(E)}\ 149}$.