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"
(Standardized answer choices, moved TOC)
Line 2: Line 2:
 
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^\text{th}</math> term of the sequence is <math>\log{b^n}</math>. What is <math>n</math>?
 
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^\text{th}</math> term of the sequence is <math>\log{b^n}</math>. What is <math>n</math>?
  
−
<math>\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 112 \qquad \textbf{(E)}\ 143</math>
+
<math>\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143</math>
  
 +
__TOC__
 
==Solution==  
 
==Solution==  
 
===Solution 1===
 
===Solution 1===

Revision as of 00:39, 26 April 2008

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

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 $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow D$.

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 $b^{13+9*11}=b^{112} \Rightarrow D$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_16&oldid=25508"