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Math Revolution and GMAT Club Contest! There is a sequence An such

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Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 09:38
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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


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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 11:12
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So, lets start by plugging in the values available to us, and see if it gives us any help solving the question. Following this approach, we can calculate A3= 5/2, A4=1/2, A5=1/5, A6=2/5 A7=2 and A8=5. We can observe a pattern in the values now. A1=A8 and A8=A2. Thus the function will result values after each 6th term. Now lets find out what term will A149 be. Divide by 6, remainder is 5. Thus its the fifth term. And we know from our previous calculations A5=1/5. Thus the answer is E.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 11:43
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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: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 14:15
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A1=2, A2=5, and An=An−1An−2,when n is an integer greater than 2.

A3= 5/2
A4=1/2
A5=1/5
A6=2/5
A7=2
A8=5
A9=5/2
A10=1/2
A11=1/5

HERE WE FIND REPEATTITION OF NUMBER AFTER EVERY SIXTH ONE.. THUS 149 WILL BE EQUAL TO A5TH TERM..
SO ANS 1/5 E
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 14:32
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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: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 16:55
1
A1 = 2
A2 = 5
A3 = 5/2
A4 = (5/2)/(5) = 1/2
A5 = (1/2)/(5/2) = 1/5
A6 = (1/5)/(1/2) = 2/5
A7 = 2 = A1
A8 = 5 = A2
This is a closed loop of 6 terms that will keep on repeating themselves
A149 = A5 = 1/5
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 17:52
The Answer is 1/2 or D.

Just quickly calculate the sequence here which turns out to be - 2, 5, 5/2, 1/2, 1/5, 2/5, 12, 2, 1/6, 1/12, 1/2, 6, 12, 2, 1/6 1/12, 2, 1/6, 1/12, 1/2, 6, 12, 2 ... (this continues with the same numbers for every 5 numbers after 2. So if you notice every 8th, 14th, 20th number is 2. an+1=a1+d*an (applying this you will find n=23 will yield a23=146). So the 146th number is 2. To find 149th number just follow the sequence of 2, 1/6, 1/12, 1/2, 6, 12. This will yield the answer 1/2.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 19:29
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\(A_1 = 2\)
\(A_2 = 5\)
\(A_3 = A_2/A_1 = 5/2\)
\(A_4 = A_3/A_2 = (5/2)/5 = 1/2\)
\(A_5 = A_4/A_3 = (1/2)/(5/2) = 1/5\)
\(A_6 = A_5/A_4 = (1/5)/(1/2) = 2/5\)
\(A_7 = A_6/A_5 = (2/5)/(1/5) = 2\) 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: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 06 Dec 2015, 22:46
1
A1 = 2
A2 = 5
A3 = A2/A1 = 5/2
A4 = A3/A2 = 1/2
A5 = A4/A3 = 1/5
A6 = A5/A4 = 2/5
A7 = A6/A5 = 2

So the sequence repeats after A6 --> A149 = A5 = 1/5

Answer: E
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 07 Dec 2015, 06:16
If u simply plug in the values you will notice a trend from A3 to A9 and after that it repeats.

Hence 149th term will be equal to 4 th term which is 1/2.

D.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 07 Dec 2015, 06:56
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OA: E

Solution
Let's see if there is any pattern
A1 = 2, A2 = 5, A3 = 5/2, A4 = 1/2, A5 = 1/5, A6 = 2/5,
A7 = 2, A8 = 5, A9 = 5/2, A10 = 1/2, A11 = 1/5, A12 = 2/5, and so on. So the cyclicity of this sequence is 6.
149 = 24 times 6 + 5. Hence, A149 = A5 = 1/5.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 07 Dec 2015, 07:36
1
Need to find the value of A149. Since this value is very advanced in the progression, we would expect to see some cycle in the sequence given.

A1=2
A2=5
A3=5/2=2.5
A4=2.5/5=0.5
A5=0.5/2.5=1/5=0.2
A6=0.2/0.5=0.4
A7=0.4/0.2=2
A8=2/0.4=5

From here we can make out that, since A7 and A8 have the same values as A1 and A2, we will see the same cycle being repeated. Hence our cyclicity is 6.

149 will have a remainder of 5 when divided by 6. So the value of A149=A5=1/5

Answer: E
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 07 Dec 2015, 08:01
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Given A 1 =2 , A 2 =5 , and A n =A n−1 / A n−2
So A 3 can be written as A 2/A 1 => 5/2
A 4 can be written as A 3/A 2 => 1/2
A 5 can be written as A 4/A 3 => 1/5
A 6 can be written as A 5/A 4 => 2/5
A 7 can be written as A 6/A 5 => 2
If you notice the pattern, the sequence becomes - 2, 3, 5/2, 1/2, 1/5, 2/5, 2, 5, ..... i.e. it repeats after 6 terms... such that every 6th term is 2/5.... hence since 150 is a multiple of 6, the 150th term of A 150 will be 2/5, which means 149th term will be 1/5.

The key to solving such questions is to determine a pattern; then its a cake walk.

Answer E.
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 07 Dec 2015, 21:48
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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: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 08 Dec 2015, 02:51
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\(A1 = 2, A2 = 5 and An = An-1/An-2\)
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: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 08 Dec 2015, 09:19
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\(A_3=5/2; A_4=1/2; A_5=1/5; A_6=2/5; A_7=2=A_1\)
\(The cycle is 6:A_1 = A_7 = A_{13} = ... = A_{144} = A_{151}\)
\(A_5=...=A_{149}=1/5\)
Answer E.
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New post 08 Dec 2015, 13:25
Answer C) 2.5

If you plug in the numbers,
A0=(2/5)
A1=2
A2=5
A3=(5/2)
A4=25/2
A5=5
A6=(2/5) [Same as A0]
A7=2 [Same as A1] so the pattern repeats it self in 7s so A149 = 7*21 +2 is the same as A3= 2.5
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 09 Dec 2015, 09:30
\(A3 =\frac{A2}{A1}\)
\(A4 =\frac{A3}{A2}= \frac{A2}{A1} *\frac{1}{A2} = \frac{1}{A1}\)
\(A5 =\frac{A4}{A3}= \frac{1}{A1}* \frac{A1}{A2}=\frac{1}{A2}\)
\(A6 = \frac{A5}{A4} = \frac{1}{A2} * A1 = \frac{A1}{A2}\)
\(A7 =\frac{A6}{A5}=\frac{A1}{A2}* A2 = A1\)
\(A8 = \frac{A7}{A6}= A1 * \frac{A2}{A1}= A2\)
\(A9 =\frac{A8}{A9}= \frac{A2}{A1}= A3\)

Therefore A7=A1 and A8 = A2 and A9 = A3

From A3 to A9, we have a cycle of 6
So, \(\frac{149}{6} = 24 \frac{5}{6}\)Remainder is 5
A149 = A(144+5) = A(3+5) = A8 = A2 = 5
Option B
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 09 Dec 2015, 10:20
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After drawing out the sequence - we see a repeating pattern :

A1 = 2 ; A2 = 5 ; A3 = 5/2 ; A4 = 1/2 ; A5 = 1/5 ; A6 = 2/5
A7 = 2 ; A 8 = 5 ; A9 = 5/2.

The pattern repeats after six consecutive terms in the series.

149 = (24*6) + 5 ; Essentially 149th term is 5th term after 24 complete cycles of 6 consecutive terms in the series.

we're looking at A5 , that is = 1/5 -- Option E
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Re: Math Revolution and GMAT Club Contest! There is a sequence An such  [#permalink]

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New post 09 Dec 2015, 11:29
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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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