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2025 AMC 10A Problems/Problem 11: Difference between revisions

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(Problem goes here)
The sequence <imath>1,x,y,z</imath> is arithmetic. The sequence <imath>1,p,q,z</imath> is geometric. Both sequences are strictly increasing and contain only integers, and <imath>z</imath> is as small as possible. What is the value of <imath>x+y+z+p+q</imath>?
 
<imath>\textbf{(A) } 66 \qquad\textbf{(B) } 91 \qquad\textbf{(C) } 103 \qquad\textbf{(D) } 132 \qquad\textbf{(E) } 149</imath>


==Solution 1==
==Solution 1==

Revision as of 14:51, 6 November 2025

The sequence $1,x,y,z$ is arithmetic. The sequence $1,p,q,z$ is geometric. Both sequences are strictly increasing and contain only integers, and $z$ is as small as possible. What is the value of $x+y+z+p+q$?

$\textbf{(A) } 66 \qquad\textbf{(B) } 91 \qquad\textbf{(C) } 103 \qquad\textbf{(D) } 132 \qquad\textbf{(E) } 149$

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

~Minor edit by dodobird150

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