2025 AMC 10A Problems/Problem 11: Difference between revisions
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==Solution 1== | ==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> | 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 <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. | A more generalized solution is as follows. | ||
Revision as of 13:33, 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. 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 
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
