Difference between revisions of "2019 AMC 10A Problems/Problem 15"
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==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) | |
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==Solution 3== | ==Solution 3== | ||
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~jakeg314 | ~jakeg314 | ||
− | ==Solution 4 ( | + | == Solution 4 (Arithmetic Sequence) == |
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− | + | 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. | |
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==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.