Difference between revisions of "2008 AMC 12A Problems/Problem 16"
(New page: ==Problem== The numbers <math>\log(a^3b^7)</math>, <math>\log(a^5b^{12})</math>, and <math>\log(a^8b^{15})</math> are the first three terms of an arithmetic sequence, and the <math>12^\te...) |
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Since the first three terms in the sequence are <math>13B</math>, <math>22B</math>, and <math>31B</math>, the <math>k</math>th term is <math>(9k + 4)B</math>. | Since the first three terms in the sequence are <math>13B</math>, <math>22B</math>, and <math>31B</math>, the <math>k</math>th term is <math>(9k + 4)B</math>. | ||
| − | Thus the <math>12^\text{th}</math> term is <math>(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow D</math> | + | Thus the <math>12^\text{th}</math> term is <math>(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow D</math>. |
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2008|ab=A|num-b=15|num-a=17}} | ||
Revision as of 14:52, 17 February 2008
Problem
The numbers 
, 
, and 
are the first three terms of an arithmetic sequence, and the 
term of the sequence is 
. What is 
?
Solution
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 
.
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 | |
