Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 12A Problems/Problem 16"

(→‎Solution)
(→‎Solution 1)
Line 14: Line 14:
 
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 \boxed{D}</math>.
  
 
===Solution 2===
 
===Solution 2===

Revision as of 15:52, 22 April 2016

Problem

The numbers $\log(a^3b^7)$, $\log(a^5b^{12})$, and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence, and the $12^\text{th}$ term of the sequence is $\log{b^n}$. What is $n$?

$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$

Solution 1

Let $A = \log(a)$ and $B = \log(b)$.

The first three terms of the arithmetic sequence are $3A + 7B$, $5A + 12B$, and $8A + 15B$, and the $12^\text{th}$ term is $nB$.

Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$.

Since the first three terms in the sequence are $13B$, $22B$, and $31B$, the $k$th term is $(9k + 4)B$.

Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}$.

Solution 2

If $\log(a^3b^7)$, $\log(a^5b^{12})$, and $\log(a^8b^{15})$ are in arithmetic progression, then $a^3b^7$, $a^5b^{12}$, and $a^8b^{15}$ are in geometric progression. Therefore,

\[a^2b^5=a^3b^3 \Rightarrow a=b^2\]

Therefore, $a^3b^7=b^{13}$, $a^5b^{12}=b^{22}$, therefore the 12th term in the sequence is $b^{13+9*11}=b^{112} \Rightarrow D$

Solution 3

$\ \text{If a, b, and c are in a arithmetic progression then } b = \frac{a+c}{2} \text{ which means}$ $\ \log(a^5b^{12}) = \frac{\log(a^3b^7) + \log(a^8b^{15})}{2}= \frac{\log(a^{11}b^{22})}{2} \text{ therefore}$ $\ 2\log(a^5b^{12}) = \log(a^{10}b^{24}) = \log(a^{11}b^{22}) \Rightarrow a=b^2$ $\ \text{This means that the Kth term of the series would be } \log(b^{13+9(k-1)})$ $\ \text{The 12th term would be } \log(b^{112}) \Rightarrow n=112 \Rightarrow D$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_16&oldid=78211"