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

Revision as of 16:04, 9 October 2022

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 $n$ ?

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

Solutions

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

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 $d$ : $d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5$ $d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.$ These two different yet equal expressions allow us to express $a$ and $b$ in terms of each other: $\log a^2b^5 = \log a^3b^3$ $a^2b^5 = a^3b^3$ $a=b^2.$ The desired $12$ th term in the sequence is $a+11d$ , so we can substitute our values for $a$ and $d$ : $a+11d = \log a^3b^7 + 11\log(a^2b^5)$ $= \log a^3b^7 + \log(a^{22}b^{55})$ $= \log a^{25}b^{62}.$ Plugging in $a=b^2$ so we can express the answer as $\log(b^n)$ , we have: $a+11d = \log b^{50}b^{62}$ $= \log b^{112} \Rightarrow \boxed{D}.$

@@ Line 29: / Line 29: @@
 <math>\ \text{This means that the Kth term of the series would be } \log(b^{13+9(k-1)})</math>
 <math>\ \text{The 12th term would be } \log(b^{112})  \Rightarrow n=112  \Rightarrow D </math>
+=== Solution 4 ===
+Given the first three terms form an arithmetic progression, we have:
+<cmath>a = \log(a^3b^7)</cmath>
+<cmath>a+d = \log(a^5b^{12})</cmath>
+<cmath>a+2d = \log(a^8b^{15}).</cmath>
+Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for <math>d</math>:
+<cmath>d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5</cmath>
+<cmath>d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.</cmath>
+These two different yet equal expressions allow us to express <math>a</math> and <math>b</math> in terms of each other:
+<cmath>\log a^2b^5 = \log a^3b^3</cmath>
+<cmath>a^2b^5 = a^3b^3</cmath>
+<cmath>a=b^2.</cmath>
+The desired <math>12</math>th term in the sequence is <math>a+11d</math>, so we can substitute our values for <math>a</math> and <math>d</math>:
+<cmath>a+11d = \log a^3b^7 + 11\log(a^2b^5)</cmath>
+<cmath> = \log a^3b^7 + \log(a^{22}b^{55})</cmath>
+<cmath> = \log a^{25}b^{62}.</cmath>
+Plugging in <math>a=b^2</math> so we can express the answer as <math>\log(b^n)</math>, we have:
+<cmath>a+11d = \log b^{50}b^{62}</cmath>
+<cmath> = \log b^{112} \Rightarrow \boxed{D}.</cmath>
 == See Also ==
 {{AMC12 box|year=2008|ab=A|num-b=15|num-a=17}}
 {{MAA Notice}}

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

Page

Toolbox

Search