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If An=An-1/An-2, A1=1, and A2=-2, then A100=?

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If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post 19 Jul 2018, 02:31
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[GMAT math practice question]

If \(A_n = \frac{A_{n-1}}{A_{n-2}}\), \(A_1=1\), and \(A_2 =-2\), then \(A_{100}=\)?


\(A. 1\)

\(B. -1\)

\(C. 2\)

\(D. -2\)

\(E. \frac{-1}{2}\)

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If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post Updated on: 21 Jul 2018, 04:09
5
Given A1=1 , A2= -2
then acc. to question we can find :
A3 = -2/1 = -2
A4= -2/-2 = 1
like wise :
A5 = -1/2
A6 = -1/2
A7= 1
A8= -2
.
.
.
now if you see after A6 cycle has started again , so for 100th term = 6*16 +4
cycle will repeat 16 times and A100 = A4

answer = 1
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Originally posted by GMATbuster92 on 19 Jul 2018, 03:52.
Last edited by GMATbuster92 on 21 Jul 2018, 04:09, edited 2 times in total.
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Re: If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post 19 Jul 2018, 02:46
1
MathRevolution wrote:
[GMAT math practice question]

If An=An-1/An-2(n3), A1=1, and A2=-2, then A100=?

\(A. 1\)
\(B. -1\)
\(C. 2\)
\(D. -2\)
\(E. \frac{-1}{2}\)


The question is not clear, MathRevolution - What does n3 mean in the question stem?
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Re: If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post 19 Jul 2018, 03:39
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Re: If An=An-1/An-2 (n3), A1=1, and A2=-2, then A100=?  [#permalink]

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New post 19 Jul 2018, 04:01
Quote:

If An=An-1/An-2 for all values n≥3, A1=1, and A2=-2, then A100=?

\(A. 1\)
\(B. -1\)
\(C. 2\)
\(D. -2\)
\(E. \frac{-1}{2}\)


\(A₁ = 1\)
\(A₂ = -2\)
\(A₃ = \frac{-2}{1} = -2\)
\(A₄ = \frac{-2}{-2} = 1\)
\(A₅ = \frac{1}{-2} = -0.5\)
\(A₆ = \frac{-0.5}{1} = -0.5\)

\(A₇ = \frac{-0.5}{-0.5} = 1\)
\(A₈ = \frac{1}{-0.5} = -2\)

As illustrated by the results in blue, A₁ through A₆ represent one complete 6-term CYCLE:
\(1\), \(-2\), \(-2\), \(1\), \(-0.5\), \(-0.5\)
Starting with A₇ and A₈, the 6-term cycle begins again:
1, -2...

Since 16*6 = 96, the 96th term represents the end of the 16th 6-term cycle.
Starting with A₉₇, the 6-term cycle begins again:
A₉₇ = 1
A₉₈ = -2
A₉₉ = -2
A₁₀₀ = 1


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Re: If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post 19 Jul 2018, 05:20

Solution



Given:

    • \(A_n = \frac{A_{n-1}}{A_{n-2}}\)
    • \(A_1= 1\)
    • \(A_2= -2\)

To find:
• \(A_{100}\).

Approach and Working:

    • \(A_3 =\frac{A_2}{A_1}= \frac{-2}{1}= -2\)
    • \(A_4= \frac{A_3}{A_2}= \frac{-2}{-2}= 1\)
    • \(A_5= \frac{A_4}{A_3}= \frac{1}{-2}= -0.5\)
    • \(A_6= \frac{A_5}{A_4}= \frac{-0.5}{1}= -0.5\)
    • \(A_7=\frac{A_6}{A_5}= \frac{-0.5}{-0.5}= 1\)
    • \(A_8= \frac{A_7}{A_6}= \frac{1}{-0.5}= -2\)
    • \(A_9=\frac{A_8}{A_7}= \frac{-2}{1}= -2\)
      o If you see \(A_1= A_7= 1\) , \(A_2= A_8=-2\), and \(A_3= A_9=-2\) then we can say that An repeats after every 6 number.
      o Hence, \(A_{100} = A_ {6*16+4}= A_4= 1\)


Hence, the correct answer is option A.

Answer: A
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Re: If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post 19 Jul 2018, 06:44
MathRevolution wrote:
[GMAT math practice question]

If \(A_n = \frac{A_{n-1}}{A_{n-2}}\), \(A_1=1\), and \(A_2 =-2\), then \(A_{100}=\)?


\(A. 1\)

\(B. -1\)

\(C. 2\)

\(D. -2\)

\(E. \frac{-1}{2}\)



\(A_n = \frac{A_{n-1}}{A_{n-2}}............A_{n-1}=A_n*A_{n-2}\)
So the middle term is multiple of the side terms..
\(A_2=A_1*A_3.........-2=1*A_3........A_3=-2\)..
\(-2=-2*A_4......A_4=1\)..
\(1=-2*A_5.......A_5=\frac{-1}{2}\)..
\(\frac{-1}{2}=1*A_6......A_6=\frac{-1}{2}\)..
\(\frac{-1}{2}=\frac{-1}{2}*A_7......A_7=1\)...
So numbers are 1,-2,-2,1,-1/2,-1/2,1,-2.....
Thus sequence gets repeated after every 6 terms..
100=96+4=6*16+4
So 4th term of the sequence....1

A
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Re: If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post 19 Jul 2018, 13:29
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
so, 2^30 the unit digit is 4
why same rule is not applicable for above question?
A0=-1/2
A1=1
.
.
.
.
A5=-1/2
A6=-1/2
A7=1
so, from A6 the value is repeating, Hence, (100/5)=20 times, means A100 must be -1/2
why it is A100=1 ???????????
kindly ans Plz.
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Re: If An=An-1/An-2, A1=1, and A2=-2, then A100=?  [#permalink]

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New post 22 Jul 2018, 18:21
=>
A1 = 1, A2 =-2.
A3 = A2 / A1 = \(\frac{(-2)}{1} = -2\).
A4 = A3 / A2 = \(\frac{(-2)}{(-2)} = 1.\)
A5 = A4 / A3 = \(\frac{1}{(-2)} = \frac{-1}{2}.\)
A6 = A5 / A4 = \(-(\frac{1}{2}) / 1 = \frac{-1}{2}.\)
A7 = A6 / A5 = \(-\frac{(1}{2)}\) / \(-(\frac{1}{2})\) = 1 = A1.
A8 = A7 / A6 = \(1 / (\frac{-1}{2}) = -2\)= A2.
Thus, every 6th term is the same. That is,
A1 = A7 = A13 = … = \(1\).
A2 = A8 = A14 = … =\(-2.\)
A3 = A9 = A15 = … = \(-2\).
A4 = A10 = A16 = … =\(1.\)
A5 = A11 = A17 = … = \(-(\frac{1}{2}).\)
A6 = A12 = A18 = … = \(-(\frac{1}{2}).\)
Now, \(100 = 6 * 16 + 4.\) Thus,
A100 = A4 = \(1\).
Therefore, the answer is A.
Answer: A
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Re: If An=An-1/An-2, A1=1, and A2=-2, then A100=?   [#permalink] 22 Jul 2018, 18:21
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