2025 AMC 10A Problems/Problem 11

(Problem goes here)

Solution 1

Since the geometric sequence is more restrictive, we can test values for the common ratio until we find one that works. The first sequence is an arithmetic sequence, so $z-1$ must be divisible by 3. After a few tests, we find that a common ratio of $4$ results in the geometric sequence $1,4,16,64,$ so the arithmetic sequence is $1,22,43,64.$ The answer is $4+16+64+22+43=\boxed{\text{(E) }149}.$

A more generalized solution is as follows. Let the common difference of the arithmetic sequence be $d$ , and the common ratio of the geometric sequence be $r.$ Hence, the two sequences are $1,1+d,1+2d,1+3d$ and $1,r,r^2,r^3.$ Since $z=1+3d=r^3,$ the arithmetic sequence is $1,1+d,1+2d,r^3.$ Since $d=\dfrac{1+3d-1}{3}=\dfrac{r^3-1}{3}$ is a positive integer, we seek the smallest $r\neq1$ such that $r^3-1=(r-1)(r^2+r+1)$ is divisble by $3,$ so the smallest $r$ is $4$ . The rest follows like above.

~Tacos_are_yummy_1

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2025 AMC 10A Problems/Problem 11

Solution 1