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Difference between revisions of "2008 AMC 12A Problems/Problem 16"

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<cmath>a^2b^5=a^3b^3 \Rightarrow a=b^2</cmath>
 
<cmath>a^2b^5=a^3b^3 \Rightarrow a=b^2</cmath>
  
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Therefore, <math>a^3b^7=b^{13}</math>, <math>a^5b^{12}=b^{22}</math>, therefore the 12th term in the sequence is <math>b^{13+9*11}=b^{112} \Rightarrow D</math>
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Therefore, <math>a^3b^7=b^{13}</math>, <math>a^5b^{12}=b^{22}</math>, therefore the 12th term in the sequence is <math>b^{13+9*11}=b^{112} \Rightarrow \boxed{D}</math>
  
 
===Solution 3===
 
===Solution 3===

Revision as of 15:53, 22 April 2016

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 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

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

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

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