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If An=An1/An2, A1=1, and A2=2, then A100=?
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19 Jul 2018, 01:31
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[GMAT math practice question] If \(A_n = \frac{A_{n1}}{A_{n2}}\), \(A_1=1\), 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=An1/An2, A1=1, and A2=2, then A100=?
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Updated on: 21 Jul 2018, 03:09
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, 02:52.
Last edited by GMATbuster92 on 21 Jul 2018, 03:09, edited 2 times in total.




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Re: If An=An1/An2, A1=1, and A2=2, then A100=?
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19 Jul 2018, 01:46
MathRevolution wrote: [GMAT math practice question]
If An=An1/An2(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=An1/An2, A1=1, and A2=2, then A100=?
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19 Jul 2018, 02:39



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Re: If An=An1/An2 (n3), A1=1, and A2=2, then A100=?
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19 Jul 2018, 03:01
Quote: If An=An1/An2 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 6term CYCLE: \(1\), \(2\), \(2\), \(1\), \(0.5\), \(0.5\) Starting with A₇ and A₈, the 6term cycle begins again: 1, 2... Since 16*6 = 96, the 96th term represents the end of the 16th 6term cycle. Starting with A₉₇, the 6term cycle begins again: A₉₇ = 1 A₉₈ = 2 A₉₉ = 2 A₁₀₀ = 1
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Re: If An=An1/An2, A1=1, and A2=2, then A100=?
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19 Jul 2018, 04:20
Solution Given:• \(A_n = \frac{A_{n1}}{A_{n2}}\) • \(A_1= 1\) • \(A_2= 2\) 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=An1/An2, A1=1, and A2=2, then A100=?
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19 Jul 2018, 05:44
MathRevolution wrote: [GMAT math practice question]
If \(A_n = \frac{A_{n1}}{A_{n2}}\), \(A_1=1\), and \(A_2 =2\), then \(A_{100}=\)?
\(A. 1\)
\(B. 1\)
\(C. 2\)
\(D. 2\)
\(E. \frac{1}{2}\) \(A_n = \frac{A_{n1}}{A_{n2}}............A_{n1}=A_n*A_{n2}\) 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=An1/An2, A1=1, and A2=2, then A100=?
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19 Jul 2018, 12:29
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=An1/An2, A1=1, and A2=2, then A100=?
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22 Jul 2018, 17:21
=> A 1 = 1, A 2 =2. A 3 = A 2 / A 1 = \(\frac{(2)}{1} = 2\). A 4 = A 3 / A 2 = \(\frac{(2)}{(2)} = 1.\) A 5 = A 4 / A 3 = \(\frac{1}{(2)} = \frac{1}{2}.\) A 6 = A 5 / A 4 = \((\frac{1}{2}) / 1 = \frac{1}{2}.\) A 7 = A 6 / A 5 = \(\frac{(1}{2)}\) / \((\frac{1}{2})\) = 1 = A 1. A 8 = A 7 / A 6 = \(1 / (\frac{1}{2}) = 2\)= A 2. Thus, every 6th term is the same. That is, A 1 = A 7 = A 13 = … = \(1\). A 2 = A 8 = A 14 = … =\(2.\) A 3 = A 9 = A 15 = … = \(2\). A 4 = A 10 = A 16 = … =\(1.\) A 5 = A 11 = A 17 = … = \((\frac{1}{2}).\) A 6 = A 12 = A 18 = … = \((\frac{1}{2}).\) Now, \(100 = 6 * 16 + 4.\) Thus, A 100 = A 4 = \(1\). Therefore, the answer is A. Answer: A
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Re: If An=An1/An2, A1=1, and A2=2, then A100=? &nbs
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