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There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),

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There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 06 Dec 2015, 08:38
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E

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Question Stats:

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Math Revolution and GMAT Club Contest Starts!



QUESTION #8:

There is a sequence \(A_n\) such that \(A_1=2\), \(A_2=5\), and \(A_n=\frac{A_{n-1}}{A_{n-2}}\),when n is an integer greater than 2. What is the value of \(A_{149}\)?

A. 2
B. 5
C. 2.5
D. 1/2
E. 1/5


Check conditions below:



Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299!



All announcements and winnings are final and no whining :-) GMAT Club reserves the rights to modify the terms of this offer at any time.


NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!


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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 06 Dec 2015, 10:43
2
this is a "spot the pattern" question type.

nothing difficult.
a3 = 2.5
a4=0.5
a5=0.2
a6=0.4
a7=2
a8=5
a9=2.5
a10=0.5
a11=0.2
a12=0.4
a13=2
a14=5

we can see that every the pattern repeats every 6th term. ex. 1+6 = 7. 7+6=13. so term a1, a7, a13, etc. will be 1/5.


now, we can deduct that term a149 will be 2.
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 06 Dec 2015, 13:32
1
We have a series defined in the question, and we are asked for the 149th term in the series.

The first thing to check is whether the series repeats. It could be expected due to the fact that the nth term depends on a division of the previous two terms, and also we know that this can often be the case in the GMAT when asked for a term much later in the series.

The series goes:

\(2\), \(5\), \(5/2\), \(1/2\), \(1/5\), \(2/5\), \(2\), \(5\)

We can see that the series has repeated, and will therefore loop forever, with 6 unique locations.

If we divide 149 by 6, the remainder will tell us which term in the series is the 149th term. 149 divided by 6 gives us the remainder 5, therefore the fifth term, \(1/5\) will also be the 149th term. Therefore the answer is E.
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 06 Dec 2015, 18:29
1
\(A_1 = 2\)
\(A_2 = 5\)
\(A_3 = A_2/A_1 = 5/2\)
\(A_4 = A_3/A_2 = (5/2)/5 = 1/2\)
\(A_5 = A_4/A_3 = (1/2)/(5/2) = 1/5\)
\(A_6 = A_5/A_4 = (1/5)/(1/2) = 2/5\)
\(A_7 = A_6/A_5 = (2/5)/(1/5) = 2\) same as \(A_1\)
\(A_8 = A_7/A_6 = (2)/(2/5) = 5\) same as \(A_2\)

So this sequence has the cyclicity of 6
\(A_{149} = A_5 = 1/5\) ( 149 when divided by 6 gives remainder of 5)

Answer (E)
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 07 Dec 2015, 20:48
1
QUESTION #8:

There is a sequence An such that A1=2, A2=5, and An=An−1/An−2,when n is an integer greater than 2. What is the value of A149?

Solution:
A1=2
A2=5
A3=A3−1/A3−2=A2/A1=5/2
A4=A3/A2=5/2*1/5=1/2
A5=A4/A3=1/2*2/5=1/5
A6=A5/A4=1/5*2=2/5
A7=A6/A5=2/5*5=2
A8=A7/A6=2*5/2=5

Hereby, the series is: A1, A2, A3, A4, A5, A6, A7, A8=2,5, 5/2,1/2, 1/5,2/5,2,5
A149=the fifth term=1/5

Answer: 'E'
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 08 Dec 2015, 01:51
1
\(A1 = 2, A2 = 5 and An = An-1/An-2\)
We get
\(A3 = 5/2\)
\(A4 = 1/2\)
\(A5 = 1/5\)
\(A6 = 2/5\)
\(A7 = 2\)
\(A8 = 5\)

We have a sequence that repeats every 6 terms
A1 = A7 and when 7 is divided by 6 the remainder is 1
A2 = A8 ... \(8/6\) = 2R

So we need to find the remainder when 149 is divided by 6 = 149/6 = 5R
So the answer is 1/5 answer E
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 09 Dec 2015, 10:29
1
Formula to find terms of sequence is An = An-1/An-2, where n>2. (first and second term are already given)

A1 = 2 (Given)
A2 = 5 (Given)
A3 = A2/A1 = 2.5 (using An = An-1/An-2)
A4 = A3/A2 = 1/2 (using An = An-1/An-2)
A5 = A4/A3 = 1/5 (using An = An-1/An-2)
A6 = A5/A4 = 2/5 (using An = An-1/An-2)
A7 = A6/A5 = 2 (using An = An-1/An-2) - A7 = A1, this is where terms has started to repeat.
A8 = A7/A6 = 5 (using An = An-1/An-2) - A8 = A2

We can generalize that there are 6 different terms in sequence and then next six terms of the sequence are same.

150 is the multiple of 6, hence A150 would be equal to sixth term of sequence i.e. 2/5.

and A149 = fifth term of sequence i.e. 1/5.

Option E is the correct answer.

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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 09 Dec 2015, 20:29
2
1
There is a sequence An such that A1=2, A2=5, and An=An−1An−2,when n is an integer greater than 2. What is the value of A149?
A1 = 2
A2 = 5
A3 = 5/2
A4= 1/2
A5 = 1/5
A6 = 2/5
A7 = 2
A8 = 5

The numbers are repeating after six steps. The closest multiple of 6 to 149 is 144 . We are 5 five short to 149 .
So A149 = A5 Which is 1/5 so the answer is E
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 14 Dec 2015, 08:13
Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!



QUESTION #8:

There is a sequence \(A_n\) such that \(A_1=2\), \(A_2=5\), and \(A_n=\frac{A_{n-1}}{A_{n-2}}\),when n is an integer greater than 2. What is the value of \(A_{149}\)?

A. 2
B. 5
C. 2.5
D. 1/2
E. 1/5


Check conditions below:



Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299!



All announcements and winnings are final and no whining :-) GMAT Club reserves the rights to modify the terms of this offer at any time.


NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!



MATH REVOLUTION OFFICIAL SOLUTION:

Generally it is best to substitute in solving sequence questions. The sequence for this question is, \(A_1=2\), \(A_2=5\), \(A_3=\frac{A_2}{A_1}=\frac{5}{2}\), \(A_4=\frac{A_3}{A_2}=(5/2)/5=\frac{1}{2}\), \(A_5=\frac{1}{5}\) and \(A_6=\frac{2}{5}\). This means \(A_{n+6}=A_n\). Then, if we divide \(149=6*24+5\) by 6, the remainder is 5. So, \(A_{149}=A_6*24+5=A_5=\frac{1}{5}\) and E is a correct answer.
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 18 Feb 2017, 00:50
Bunuel wrote:
Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!



QUESTION #8:

There is a sequence \(A_n\) such that \(A_1=2\), \(A_2=5\), and \(A_n=\frac{A_{n-1}}{A_{n-2}}\),when n is an integer greater than 2. What is the value of \(A_{149}\)?

A. 2
B. 5
C. 2.5
D. 1/2
E. 1/5


Check conditions below:



Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299!



All announcements and winnings are final and no whining :-) GMAT Club reserves the rights to modify the terms of this offer at any time.


NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!



MATH REVOLUTION OFFICIAL SOLUTION:

Generally it is best to substitute in solving sequence questions. The sequence for this question is, \(A_1=2\), \(A_2=5\), \(A_3=\frac{A_2}{A_1}=\frac{5}{2}\), \(A_4=\frac{A_3}{A_2}=(5/2)/5=\frac{1}{2}\), \(A_5=\frac{1}{5}\) and \(A_6=\frac{2}{5}\). This means \(A_{n+6}=A_n\). Then, if we divide \(149=6*24+5\) by 6, the remainder is 5. So, \(A_{149}=A_6*24+5=A_5=\frac{1}{5}\) and E is a correct answer.


Dear Bunuel,

\(A_{149}=A_6*24+5=A_5=\frac{1}{5}\)

Does we conclude \(A_{149}=A_{5}\) by obtaining the remainder equal to \(5\)?
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 18 Feb 2017, 00:57
ziyuenlau wrote:
Bunuel wrote:
Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!



QUESTION #8:

There is a sequence \(A_n\) such that \(A_1=2\), \(A_2=5\), and \(A_n=\frac{A_{n-1}}{A_{n-2}}\),when n is an integer greater than 2. What is the value of \(A_{149}\)?

A. 2
B. 5
C. 2.5
D. 1/2
E. 1/5


Check conditions below:



Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299!



All announcements and winnings are final and no whining :-) GMAT Club reserves the rights to modify the terms of this offer at any time.


NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!



MATH REVOLUTION OFFICIAL SOLUTION:

Generally it is best to substitute in solving sequence questions. The sequence for this question is, \(A_1=2\), \(A_2=5\), \(A_3=\frac{A_2}{A_1}=\frac{5}{2}\), \(A_4=\frac{A_3}{A_2}=(5/2)/5=\frac{1}{2}\), \(A_5=\frac{1}{5}\) and \(A_6=\frac{2}{5}\). This means \(A_{n+6}=A_n\). Then, if we divide \(149=6*24+5\) by 6, the remainder is 5. So, \(A_{149}=A_6*24+5=A_5=\frac{1}{5}\) and E is a correct answer.


Dear Bunuel,

\(A_{149}=A_6*24+5=A_5=\frac{1}{5}\)

Does we conclude \(A_{149}=A_{5}\) by obtaining the remainder equal to \(5\)?


The sequence goes in blocks of 6 {2, 5, 5/2, 1/2, 1/5, 2/5} {2, 5, 5/2, 1/2, 1/5, 2/5} {2, 5, 5/2, 1/2, 1/5, 2/5}...

149 is a multiple of 6 (144) plus 5, thus A149 equals to 5th number in the pattern, which is 1/5.
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),  [#permalink]

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New post 01 Jul 2019, 21:23
Bunuel wrote:

Math Revolution and GMAT Club Contest Starts!



QUESTION #8:

There is a sequence \(A_n\) such that \(A_1=2\), \(A_2=5\), and \(A_n=\frac{A_{n-1}}{A_{n-2}}\),when n is an integer greater than 2. What is the value of \(A_{149}\)?

A. 2
B. 5
C. 2.5
D. 1/2
E. 1/5


Check conditions below:



Math Revolution and GMAT Club Contest

The Contest Starts November 28th in Quant Forum


We are happy to announce a Math Revolution and GMAT Club Contest

For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday).

To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.

PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:

PS + DS course with 502 videos that is worth $299!



All announcements and winnings are final and no whining :-) GMAT Club reserves the rights to modify the terms of this offer at any time.


NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.

Thank you!



A1 =2
A2 = 5
A3 =5/2
A4 = 1/2
A5 = 1/5
A6 = 2/5
A7 = 2
A8 = 5

We see that A7 = A1 & A8 = A2, in general A(6+k) = Ak
149 = 6*24 + 5
A149 = A5 = 1/5

IMO E
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Re: There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),   [#permalink] 01 Jul 2019, 21:23

There is a sequence An such A1 = 2, A2 = 5, and An = A(n-1)/A(n-2),

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