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. 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> | 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 <imath>z-1</imath> 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 
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
