Difference between revisions of "2019 AMC 10A Problems/Problem 15"

Revision as of 00:56, 5 January 2020

The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.

Problem

A sequence of numbers is defined recursively by $a_1 = 1$ , $a_2 = \frac{3}{7}$ , and $a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}$ for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

$\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078$

Solution 1 (Induction)

Using the recursive formula, we find $a_3=\frac{3}{11}$ , $a_4=\frac{3}{15}$ , and so on. It appears that $a_n=\frac{3}{4n-1}$ , for all $n$ . Setting $n=2019$ , we find $a_{2019}=\frac{3}{8075}$ , so the answer is $\boxed{\textbf{(E) }8078}$ .

To prove this formula, we use induction. We are given that $a_1=1$ and $a_2=\frac{3}{7}$ , which satisfy our formula. Now assume the formula holds true for all $n\le m$ for some positive integer $m$ . By our assumption, $a_{m-1}=\frac{3}{4m-5}$ and $a_m=\frac{3}{4m-1}$ . Using the recursive formula, $a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},$ so our induction is complete.

Solution 2

Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by $b_n = \frac{1}{a_n}$ .

We have $\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}$ , so in other words, $b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=4b_{n-3}-3b_{n-4}=\dots$ .

By recursively following this pattern, we can see that $b_n=(n-1) \cdot b_2 - (n-2) \cdot b_1$ .

By plugging in 2019, we thus find $b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8075}{3}$ . Since the numerator and the denominator are relatively prime, the answer is $\boxed{\textbf{(E) } 8078}$ .

-eric2020

Solution 3

It seems reasonable to transform the equation into something else. Let $a_{n}=x$ , $a_{n-1}=y$ , and $a_{n-2}=z$ . Therefore, we have $x=\frac{zy}{2z-y}$ $2xz-xy=zy$ $2xz=y(x+z)$ $y=\frac{2xz}{x+z}$ Thus, $y$ is the harmonic mean of $x$ and $z$ . This implies $a_{n}$ is a harmonic sequence or equivalently $b_{n}=\frac{1}{a_{n}}$ is arithmetic. Now, we have $b_{1}=1$ , $b_{2}=\frac{7}{3}$ , $b_{3}=\frac{11}{3}$ , and so on. Since the common difference is $\frac{4}{3}$ , we can express $b_{n}$ explicitly as $b_{n}=\frac{4}{3}(n-1)+1$ . This gives $b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}$ which implies $a_{2019}=\frac{3}{8075}=\frac{p}{q}$ . $p+q=\boxed{\textbf{(E) } 8078}$ ~jakeg314

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 10 Problems and Solutions

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 <math>\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078</math>
-==Solution 1==
+==Solution 1 (Induction)==
 Using the recursive formula, we find <math>a_3=\frac{3}{11}</math>, <math>a_4=\frac{3}{15}</math>, and so on. It appears that <math>a_n=\frac{3}{4n-1}</math>, for all <math>n</math>. Setting <math>n=2019</math>, we find <math>a_{2019}=\frac{3}{8075}</math>, so the answer is <math>\boxed{\textbf{(E) }8078}</math>.

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