Difference between revisions of "2008 AMC 12A Problems/Problem 16"
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<math>\ \log(a^5b^{12}) = \frac{\log(a^3b^7) + \log(a^8b^{15})}{2}= \frac{\log(a^{11}b^{22})}{2} \text{ therefore}</math> | <math>\ \log(a^5b^{12}) = \frac{\log(a^3b^7) + \log(a^8b^{15})}{2}= \frac{\log(a^{11}b^{22})}{2} \text{ therefore}</math> | ||
<math>\ 2\log(a^5b^{12}) = \log(a^{10}b^{24}) = \log(a^{11}b^{22}) \Rightarrow a=b^2</math> | <math>\ 2\log(a^5b^{12}) = \log(a^{10}b^{24}) = \log(a^{11}b^{22}) \Rightarrow a=b^2</math> | ||
− | <math>\ \text{This means that the Kth of the series would be } \log(b^{13+9(k-1)})</math> | + | <math>\ \text{This means that the Kth term of the series would be } \log(b^{13+9(k-1)})</math> |
<math>\ \text{The 12th term would be } \log(b^{112}) \Rightarrow n=112 \Rightarrow D </math> | <math>\ \text{The 12th term would be } \log(b^{112}) \Rightarrow n=112 \Rightarrow D </math> | ||
Revision as of 18:34, 28 March 2016
Problem
The numbers , , and are the first three terms of an arithmetic sequence, and the term of the sequence is . What is ?
Solution
Solution 1
Let and .
The first three terms of the arithmetic sequence are , , and , and the term is .
Thus, .
Since the first three terms in the sequence are , , and , the th term is .
Thus the term is .
Solution 2
If , , and are in arithmetic progression, then , , and are in geometric progression. Therefore,
Therefore, , , therefore the 12th term in the sequence is
Solution 3
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.