Difference between revisions of "2019 AMC 10A Problems/Problem 15"
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A sequence of numbers is defined recursively by <math>a_1 = 1</math>, <math>a_2 = \frac{3}{7}</math>, and | A sequence of numbers is defined recursively by <math>a_1 = 1</math>, <math>a_2 = \frac{3}{7}</math>, and | ||
− | <cmath>a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}</cmath>for all <math>n \geq 3</math> Then <math>a_{2019}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive | + | <cmath>a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}</cmath>for all <math>n \geq 3</math> Then <math>a_{2019}</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q ?</math> |
<math>\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078</math> | <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>. | 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>. | ||
− | + | To prove this formula, we use induction. We are given that <math>a_1=1</math> and <math>a_2=\frac{3}{7}</math>, which satisfy our formula. Now assume the formula holds true for all <math>n\le m</math> for some positive integer <math>m</math>. By our assumption, <math>a_{m-1}=\frac{3}{4m-5}</math> and <math>a_m=\frac{3}{4m-1}</math>. Using the recursive formula, <cmath>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},</cmath> | |
− | To prove this formula, we use induction. We are given that <math>a_1=1</math> and <math>a_2=\frac{3}{7}</math>, which satisfy our formula. Now assume the formula holds true for all <math>n\le m</math> for some positive integer <math>m</math>. By our assumption, <math>a_{m-1}=\frac{3}{4m-5}</math> and <math>a_m=\frac{3}{4m-1}</math>. Using the recursive formula, <cmath>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{(\frac{3}{4m-5}\cdot\frac{3}{4m-1})(4m-5)(4m-1)}{(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1})(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},</cmath> | ||
so our induction is complete. | so our induction is complete. | ||
==Solution 2== | ==Solution 2== | ||
− | Since | + | We have <math>\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}}</math>, in other words, <math>\frac{1}{a_n}-\frac{1}{a_{n-1}} = \frac{1}{a_{n-1}}-\frac{1}{a_{n-2}}</math>. So <math>\{\frac{1}{a_n}\}</math> is an arithmetic sequence with step size <math>\frac{7}{3}-1=\frac{4}{3}</math>, which means <math>\frac{1}{a_{2019}} = 1+2018 \cdot \frac{4}{3} = \frac{8075}{3}</math>. Since the numerator and the denominator are relatively prime, the answer is <math>\boxed{\textbf{(E) } 8078}</math>. |
+ | |||
+ | -eric2020 (modified by Dolphindesigner) | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | It seems reasonable to transform the equation into something else. Let <math>a_{n}=x</math>, <math>a_{n-1}=y</math>, and <math>a_{n-2}=z</math>. Therefore, we have <cmath>x=\frac{zy}{2z-y}</cmath> | ||
+ | <cmath>2xz-xy=zy</cmath> | ||
+ | <cmath>2xz=y(x+z)</cmath> | ||
+ | <cmath>y=\frac{2xz}{x+z}</cmath> | ||
+ | Thus, <math>y</math> is the harmonic mean of <math>x</math> and <math>z</math>. This implies <math>a_{n}</math> is a harmonic sequence or equivalently <math>b_{n}=\frac{1}{a_{n}}</math> is arithmetic. Now, we have <math>b_{1}=1</math>, <math>b_{2}=\frac{7}{3}</math>, <math>b_{3}=\frac{11}{3}</math>, and so on. Since the common difference is <math>\frac{4}{3}</math>, we can express <math>b_{n}</math> explicitly as <math>b_{n}=\frac{4}{3}(n-1)+1</math>. This gives <math>b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}</math> which implies <math>a_{2019}=\frac{3}{8075}=\frac{p}{q}</math>. <math>p+q=\boxed{\textbf{(E) } 8078}</math> | ||
+ | ~jakeg314 | ||
− | + | == Solution 4 (Arithmetic Sequence) == | |
− | < | + | Notice that<cmath>a_n = \frac{1}{\frac{2}{a_{n-1}} - \frac{1}{a_{n-2}}}.</cmath>Therefore,<cmath>\frac{1}{a_n} = \frac{2}{a_{n-1}} - \frac{1}{a_{n-2}}, \ \ \implies \ \ \frac{\frac{1}{a_n} + \frac{1}{a_{n-2}}}{2} = \frac{1}{a_{n-1}}.</cmath>Therefore, the sequence <math>b_n = \frac{1}{a_n}</math> is an arithmetic sequence. Notice that the common difference of <math>b</math> is <math>\frac{4}{3},</math> and therefore<cmath>b_{2019} = b_1 + 2018 \bigg(\frac{5}{3}\bigg) = 1 + 2018 \bigg(\frac{4}{3} \bigg) = \frac{8075}{3}.</cmath>Therefore, we see that <math>a_{2019} = \frac{3}{8075},</math> so that <math>p + q = \boxed{\text{(E) } 8078}.</math> |
− | + | ~Professor-Mom | |
− | + | Note: This is similar to solutions #2 and #3, although you can notice that in #2's case the new sequence <math>B</math> actually forms an arithmetic sequence. | |
==See Also== | ==See Also== |
Revision as of 10:28, 23 July 2024
- The following problem is from both the 2019 AMC 10A #15 and 2019 AMC 12A #9, so both problems redirect to this page.
Contents
Problem
A sequence of numbers is defined recursively by ,
, and
for all
Then
can be written as
, where
and
are relatively prime positive integers. What is
Solution 1 (Induction)
Using the recursive formula, we find ,
, and so on. It appears that
, for all
. Setting
, we find
, so the answer is
.
To prove this formula, we use induction. We are given that and
, which satisfy our formula. Now assume the formula holds true for all
for some positive integer
. By our assumption,
and
. Using the recursive formula,
so our induction is complete.
Solution 2
We have , in other words,
. So
is an arithmetic sequence with step size
, which means
. Since the numerator and the denominator are relatively prime, the answer is
.
-eric2020 (modified by Dolphindesigner)
Solution 3
It seems reasonable to transform the equation into something else. Let ,
, and
. Therefore, we have
Thus,
is the harmonic mean of
and
. This implies
is a harmonic sequence or equivalently
is arithmetic. Now, we have
,
,
, and so on. Since the common difference is
, we can express
explicitly as
. This gives
which implies
.
~jakeg314
Solution 4 (Arithmetic Sequence)
Notice thatTherefore,
Therefore, the sequence
is an arithmetic sequence. Notice that the common difference of
is
and therefore
Therefore, we see that
so that
~Professor-Mom
Note: This is similar to solutions #2 and #3, although you can notice that in #2's case the new sequence actually forms an arithmetic sequence.
See Also
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.