2008 AMC 12A Problems/Problem 16

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

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

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2008 AMC 12A Problems/Problem 16

Contents

Problem

Solution

Solution 1

Solution 2

See Also