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

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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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mvictor · Answer

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.

arthearoth · Answer

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.

pluto82 · Answer

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)

shapla · Answer

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'

Icecream87 · Answer

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

HKD1710 · Answer

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.

hari1985 · Answer

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

Bunuel · Answer

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 .

BillyZ · 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 ?

Bunuel · Answer

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.

Kinshook · Answer

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

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