2025 AMC 10A Problems/Problem 11: Difference between revisions
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(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. After a few tests, we find that a common ratio of <imath>4</imath> results in the geometric sequence <imath>1,4,16,64,</imath> so the arithmetic sequence is <imath>1,22,43,64.</imath> The answer is <imath>4+16+64+22+43=\boxed{\text{(E) }149}.</imath> | |||
A more generalized solution is as follows. | |||
Let the common difference of the arithmetic sequence be <imath>d</imath>, and the common ratio of the geometric sequence be <imath>r.</imath> Hence, the two sequences are <imath>1,1+d,1+2d,1+3d</imath> and <imath>1,r,r^2,r^3.</imath> Since <imath>z=1+3d=r^3,</imath> the arithmetic sequence is <imath>1,1+d,1+2d,r^3.</imath> Since <imath>d=\dfrac{1+3d-1}{3}=\dfrac{r^3-1}{3}</imath> is a positive integer, we seek the smallest <imath>r\neq1</imath> such that <imath>r^3-1=(r-1)(r^2+r+1)</imath> is divisble by <imath>3,</imath> so the smallest <imath>r</imath> is <imath>4</imath>. The rest follows like above. | |||
~Tacos_are_yummy_1 | |||
Revision as of 12:49, 6 November 2025
(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. After a few tests, we find that a common ratio of 
results in the geometric sequence 
so the arithmetic sequence is 
The answer is 
A more generalized solution is as follows.
Let the common difference of the arithmetic sequence be 
, and the common ratio of the geometric sequence be 
Hence, the two sequences are 
and 
Since 
the arithmetic sequence is 
Since 
is a positive integer, we seek the smallest 
such that 
is divisble by 
so the smallest 
is . The rest follows like above.
~Tacos_are_yummy_1
