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For the sequence a1, a2 … an, every term after the first can be found

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Math Expert
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Joined: 02 Sep 2009
Posts: 59561
For the sequence a1, a2 … an, every term after the first can be found  [#permalink]

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New post 10 Aug 2018, 03:17
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E

Difficulty:

  55% (hard)

Question Stats:

60% (02:00) correct 40% (02:44) wrong based on 52 sessions

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For the sequence \(a_1\), \(a_2\), …, an, every term after the first can be found using the equation \(a_n=2(a_{n−1})\). If \(a_1=\frac{1}{2}\), then \(a_{49}\) is what percent less than \(a_{52}\)?

A. 12.5
B. 25
C. 50
D. 75
E. 87.5
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Re: For the sequence a1, a2 … an, every term after the first can be found  [#permalink]

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New post 10 Aug 2018, 03:46
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Bunuel wrote:
For the sequence \(a_1\), \(a_2\), …, an, every term after the first can be found using the equation \(a_n=2(a_{n−1})\). If \(a_1=\frac{1}{2}\), then \(a_{49}\) is what percent less than \(a_{52}\)?

An = 2*(An-1)
A1 = 1/2
A2 = 2 * A1 = 2 * 1/2 = 1
A3 = 2 * A2 = 2 * 1 = 2
A4 = 2 * A3 = 2 * 2 = 4

Now, we are asked to find difference in % of A49 and A52
Since, it is an ongoing sequence with fixed increment every time, we can calculate the % change by considering any element(x, suppose) and third element to x in the sequence, and % difference will still be the same. Third element because A49 is 3 less than A52.

Therefore,
% difference = (( A52 - A49 ) / A52 ) * 100 = (( A4 - A1 ) / A4 ) * 100 = (( 4 - 1/2 ) / A4 ) * 100 = ( 3.5 / 4 ) * 100 = 0.875 * 100 = 87.5

Hence, E.
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For the sequence a1, a2 … an, every term after the first can be found  [#permalink]

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New post 10 Aug 2018, 05:13
Bunuel wrote:
For the sequence \(a_1\), \(a_2\), …, an, every term after the first can be found using the equation \(a_n=2(a_{n−1})\). If \(a_1=\frac{1}{2}\), then \(a_{49}\) is what percent less than \(a_{52}\)?

A. 12.5
B. 25
C. 50
D. 75
E. 87.5


Given, \(a_n=2(a_{n−1})\)
So, \(a_{52}\)=2*\(a_{51}\)=\(2^2*a_{50}\)=\(2^3*a_{49}\)

% change from \(a_{49}\) to \(a_{52}\)=\(\frac{a_{49}-a_{52}}{a_{52}}*100\)=\(\frac{a_{49}-2^3*a_{49}}{{2^3*a_{49}}}*100\)=\(\frac{1-8}{8}*100\)=-87.5%

So, \(a_{49}\) is 87.5% less than \(a_{52}\).

Ans. (E)
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For the sequence a1, a2 … an, every term after the first can be found   [#permalink] 10 Aug 2018, 05:13
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