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

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Math Expert
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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10 Aug 2018, 03:17
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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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Concentration: Operations, Marketing
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Re: For the sequence a1, a2 … an, every term after the first can be found  [#permalink]

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10 Aug 2018, 03:46
1
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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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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