2008 AMC 12A Problems/Problem 16: Difference between revisions

Revision as of 14:52, 17 February 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 $n$ ?

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

Solution

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

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

@@ Line 13: / Line 13: @@
 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}}

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2008 AMC 12A Problems/Problem 16: Difference between revisions

Revision as of 14:52, 17 February 2008

Problem

Solution

See Also