AoPSWiki

Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!

Personal tools

Search

Toolbox

2008 AMC 12A Problems/Problem 16

From AoPSWiki

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 $n$ ?

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

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See Also

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 • 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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_12A_Problems/Problem_16"

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us