2008 AMC 12A Problems/Problem 16


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

Solution 3

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

Solution 4 (mimimal manipulation)

Given the first three terms form an arithmetic progression, we have: \[a = \log(a^3b^7)\] \[a+d = \log(a^5b^{12})\] \[a+2d = \log(a^8b^{15}).\] Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for $d$: \[d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5\] \[d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.\] The desired $12$th term in the sequence is $a+11d$, so we can substitute our values for $a$ and $d$ (using either one of our two expressions for $d$): \[a+11d = \log a^3b^7 + 11\log(a^2b^5)\] \[= \log a^3b^7 + \log(a^{22}b^{55})\] \[= \log a^{25}b^{62}.\] The answer must be expressed as $\log(b^n)$, however. We're in luck: the two different yet equal expressions for $d$ allow us to express $a$ and $b$ in terms of each other: \[\log a^2b^5 = \log a^3b^3\] \[a^2b^5 = a^3b^3\] \[a=b^2.\] Plugging in $a=b^2$, we have: \[a+11d = \log b^{50}b^{62}\] \[= \log b^{112} \Rightarrow \boxed{D}.\] ~ Jingwei325 $\smiley$

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions

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