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14 May 2019, 23:38
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
44% (03:25) correct
56%
(03:00)
wrong
based on 86
sessions
History
Date
Time
Result
Not Attempted Yet
A sequence of numbers is defined recursively by \(a_1 = 1\), \(a_2 = \frac{3}{7}\), and \(a_n=\frac{a_{n-2}*a_{n-1}}{2a_{n-2} - a_{n-1}}\) 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 \(p+q\) ?
(A) 2020
(B) 4039
(C) 6057
(D) 6061
(E) 8078
(A) 2020
(B) 4039
(C) 6057
(D) 6061
(E) 8078
09 Nov 2021, 09:18
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\(a_1\)= 1
\(a_2\)= \(\frac{3}{7}\)
Using the formula given for numbers 3 and above we get,
\(a_3\)= \(\frac{3}{11}\)
\(a_4\)= \(\frac{1}{5}\)
\(a_5\)= \(\frac{3}{19}\)
\(a_6\)= \(\frac{3}{23}\)
This sequence is neither an Arithmetic Progression nor a Geometric Progression. So, let's try Harmonic Progression by taking reciprocal of all numbers.
Now we get 1, \(\frac{7}{3}\) , \(\frac{11}{3}\) , 5 , \(\frac{19}{3}\) , \(\frac{23}{3}\)
Clearly, there is a sequence, and it is an arithmetic Progression with a=1 and d=\(\frac{4}{3}\). Therefore, we can confirm that the original sequence is an Harmonic Progression.
(Note: "5" is nothing but \(\frac{15}{3}\))
In Harmonic Progression, \(a_n\) = \(\frac{ 1 }{ (a1 + (n-1)*d)}\)
\(a_{2019}\) =\(\frac{1 }{ (1 + 2018 * 4/3)}\)
1 + ( 2018 * \(\frac{4}{3}\)) = 1 + \(\frac{8072 }{ 3}\) = \(\frac{8075}{3}\)
Therefore \(a_{2019}\) = \(\frac{3}{8075}\)
(we can also confirm that the numerator p and denominator q are relatively prime by prime factorising 8075)
So, p+q = 8078
Answer is E.
Hope this helps.
---
Please give kudos if you like the simplicity of this solution
\(a_2\)= \(\frac{3}{7}\)
Using the formula given for numbers 3 and above we get,
\(a_3\)= \(\frac{3}{11}\)
\(a_4\)= \(\frac{1}{5}\)
\(a_5\)= \(\frac{3}{19}\)
\(a_6\)= \(\frac{3}{23}\)
This sequence is neither an Arithmetic Progression nor a Geometric Progression. So, let's try Harmonic Progression by taking reciprocal of all numbers.
Now we get 1, \(\frac{7}{3}\) , \(\frac{11}{3}\) , 5 , \(\frac{19}{3}\) , \(\frac{23}{3}\)
Clearly, there is a sequence, and it is an arithmetic Progression with a=1 and d=\(\frac{4}{3}\). Therefore, we can confirm that the original sequence is an Harmonic Progression.
(Note: "5" is nothing but \(\frac{15}{3}\))
In Harmonic Progression, \(a_n\) = \(\frac{ 1 }{ (a1 + (n-1)*d)}\)
\(a_{2019}\) =\(\frac{1 }{ (1 + 2018 * 4/3)}\)
1 + ( 2018 * \(\frac{4}{3}\)) = 1 + \(\frac{8072 }{ 3}\) = \(\frac{8075}{3}\)
Therefore \(a_{2019}\) = \(\frac{3}{8075}\)
(we can also confirm that the numerator p and denominator q are relatively prime by prime factorising 8075)
So, p+q = 8078
Answer is E.
Hope this helps.
---
Please give kudos if you like the simplicity of this solution
General Discussion
17 May 2019, 01:47
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This is an excellent question which tests you on your ability to observe and develop patterns in solving the problem. It is also very easy to fall into the trap of calculating the first few values of the sequence in trying to develop a pattern, which may actually complicate things more.
Instead, a good idea is to work on simplifying the expression given for \(a_n\) and to see if it has any give away, in terms of a pattern. Also, there is a piece of information in the question stating that p and q are relatively prime positive numbers. How and where you make use of this information will also have a say in the solution.
Okay, so we know that \(a_1\)= 1 and \(a_2\) = \(\frac{3}{7}\). We can’t make much out of this yet. Let us look at the expression given for \(a_n\).
\(a_n\) = \(\frac{a_{n-2} * a_{n-1}}{2a_{n-2} – a_{n-1}}\).
Is there any pattern you can observe here? No? Okay, let’s take the reciprocal of \(a_n\).
\(\frac{1}{a_n} = \frac{2a_{n-2} – a_{n-1}}{a_{n-2} * a_{n-1}}\), which can also be written as
\(\frac{1}{a_n} =\frac{2}{a_{n-1}} – \frac{1}{a_{n-2}}\).
Let us use this as a basis to calculate a few more terms of the sequence.
\(\frac{1}{a_3}\) = \(\frac{2}{a_2}\) – \(\frac{1}{a_1}\)
\(\frac{1}{a_4}\) = \(\frac{2}{a_3}\) – \(\frac{1}{a_2.}\) On substituting the value of \(\frac{1}{a_3}\)
from the previous step, this simplifies to
\(\frac{1}{a_4}\) = \(\frac{3}{a_2}\) – \(\frac{2}{a_1}\).
In a similar manner, we can determine that \(\frac{1}{a_5}\)= \(\frac{4}{a_2}\) – \(\frac{3}{a_1}\).
You can see that the two numerators on the RHS of the equation are, respectively 1 less and 2 less than the term number on the LHS of the equation. For example, for term 4, the numerators are 3 and 2; for term 5, the numerators are 4 and 3. This is clearly a pattern.
From this, we can definitely say that
\(\frac{1}{a_{2019}}\) = \(\frac{2018}{a_2}\) – \(\frac{2017}{a_1}\)
On substituting values of \(a_2\) and \(a_1\) and simplifying, we get,
\(\frac{1}{a_{2019}}\) =\(\frac{8075}{3}\). Therefore, \(a_{2019}\) = \(\frac{3}{8075}\).
Now the only task ahead of you is to confirm if 8075 is relatively prime with 3. The simplest way to do that is to prime factorise 8075. On prime factorizing, you will find that,
8075 = \(5^2\) * 17 * 19.
This means, 8075 and 3 do not have any other common factor between them and therefore, are co-prime.
So, the value of p + q is equal to 8078. The correct answer option is E.
The question giving you the values of \(a_1\) and \(a_2\) only, and then giving you an expression for an, followed by asking you to find out the value of a large term only points in one direction – this large term of yours will finally come out in terms of \(a_1\) and \(a_2\). If you are able to think in this direction, it will set the ball rolling for you to think of simplifying the given expression and developing a pattern.
You may stumble in the initial phase by going the value way, but eventually, if you are able to figure out that you have to analyse the expression to get the answer, you will end up hitting the bull’s eye.
Hope this helps!
Instead, a good idea is to work on simplifying the expression given for \(a_n\) and to see if it has any give away, in terms of a pattern. Also, there is a piece of information in the question stating that p and q are relatively prime positive numbers. How and where you make use of this information will also have a say in the solution.
Okay, so we know that \(a_1\)= 1 and \(a_2\) = \(\frac{3}{7}\). We can’t make much out of this yet. Let us look at the expression given for \(a_n\).
\(a_n\) = \(\frac{a_{n-2} * a_{n-1}}{2a_{n-2} – a_{n-1}}\).
Is there any pattern you can observe here? No? Okay, let’s take the reciprocal of \(a_n\).
\(\frac{1}{a_n} = \frac{2a_{n-2} – a_{n-1}}{a_{n-2} * a_{n-1}}\), which can also be written as
\(\frac{1}{a_n} =\frac{2}{a_{n-1}} – \frac{1}{a_{n-2}}\).
Let us use this as a basis to calculate a few more terms of the sequence.
\(\frac{1}{a_3}\) = \(\frac{2}{a_2}\) – \(\frac{1}{a_1}\)
\(\frac{1}{a_4}\) = \(\frac{2}{a_3}\) – \(\frac{1}{a_2.}\) On substituting the value of \(\frac{1}{a_3}\)
from the previous step, this simplifies to
\(\frac{1}{a_4}\) = \(\frac{3}{a_2}\) – \(\frac{2}{a_1}\).
In a similar manner, we can determine that \(\frac{1}{a_5}\)= \(\frac{4}{a_2}\) – \(\frac{3}{a_1}\).
You can see that the two numerators on the RHS of the equation are, respectively 1 less and 2 less than the term number on the LHS of the equation. For example, for term 4, the numerators are 3 and 2; for term 5, the numerators are 4 and 3. This is clearly a pattern.
From this, we can definitely say that
\(\frac{1}{a_{2019}}\) = \(\frac{2018}{a_2}\) – \(\frac{2017}{a_1}\)
On substituting values of \(a_2\) and \(a_1\) and simplifying, we get,
\(\frac{1}{a_{2019}}\) =\(\frac{8075}{3}\). Therefore, \(a_{2019}\) = \(\frac{3}{8075}\).
Now the only task ahead of you is to confirm if 8075 is relatively prime with 3. The simplest way to do that is to prime factorise 8075. On prime factorizing, you will find that,
8075 = \(5^2\) * 17 * 19.
This means, 8075 and 3 do not have any other common factor between them and therefore, are co-prime.
So, the value of p + q is equal to 8078. The correct answer option is E.
The question giving you the values of \(a_1\) and \(a_2\) only, and then giving you an expression for an, followed by asking you to find out the value of a large term only points in one direction – this large term of yours will finally come out in terms of \(a_1\) and \(a_2\). If you are able to think in this direction, it will set the ball rolling for you to think of simplifying the given expression and developing a pattern.
You may stumble in the initial phase by going the value way, but eventually, if you are able to figure out that you have to analyse the expression to get the answer, you will end up hitting the bull’s eye.
Hope this helps!
