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# There is a sequence A(n) where n is a positive integer such that A(n+1

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Joined: 05 Jun 2016
Posts: 18
GMAT 1: 760 Q51 V41
There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

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Updated on: 15 Jan 2017, 01:32
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Difficulty:

85% (hard)

Question Stats:

47% (02:33) correct 53% (02:42) wrong based on 74 sessions

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There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50

Originally posted by hwang327 on 14 Jan 2017, 14:21.
Last edited by Bunuel on 15 Jan 2017, 01:32, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

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15 Jan 2017, 01:00
2
1
hwang327 wrote:
There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50

Any explanations would be great.

Source: MathRevolution

Hi

This is combination of geometric and arithmetic sequences.

n>0, min n = 1.

$$a_2 = 10 + \frac{1}{2}a_1$$

$$a_3 = 10 + \frac{1}{2}a_2 = 10 + \frac{1}{2}(10 + \frac{1}{2}a_1) = 10 + 5 + \frac{1}{4}a_1$$

$$a_4 = 10 + \frac{1}{2}a_3 = 10 + \frac{1}{2}(10 + 5 + \frac{1}{4}a_1) = 10 + 5 + \frac{5}{2} + \frac{1}{8}a_1$$

$$a_5 = 10 + \frac{1}{2}a_4 = 10 + 5 + \frac{5}{2} + \frac{5}{4} + \frac{1}{16}a_1$$

...

$$a_n = 10 + 5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{n-3}} + \frac{1}{2}^{n-1}a_1$$

When n=1000 our fraction $$\frac{1}{2^{999}}$$ is close to 0.

$$5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{997}}$$ We can apply logic of infinite geometric series where |r|<1 because our n is quite big (1000).

$$S = \frac{5}{1-1/2} = 5*2 = 10$$

$$a_{1000} ≈ 10 + 10 + 0 = 20$$

Hope this helps

Regards
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Posts: 6961
There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

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15 Jan 2017, 04:49
2
1
hwang327 wrote:
There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50

Hi,

A point before the solution..
The Q is flawed in that there is no value of $$A_1$$ given.
Solution..
$$A_1=10, A_2=10+0.5*10=10+5, A_3=10+5+0.5*5=10+5+0.25=10+10/2+10/4+.....$$
So 1000 can be taken as infinite series..
Ans =$$\frac{a}{(1-r)}=10/(1-1/2)=10/(1/2)=20$$
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Joined: 30 Jan 2016
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Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

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16 Jan 2017, 17:03
chetan2u wrote:
hwang327 wrote:
There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?

A. 15
B. 18
C. 20
D. 25
E. 50

Hi,

A point before the solution..
The Q is flawed in that there is no value of $$A_1$$ given.
Solution..
$$A_1=10, A_2=10+0.5*10=10+5, A_3=10+5+0.5*5=10+5+0.25=10+10/2+10/4+.....$$
So 1000 can be taken as infinite series..
Ans =$$\frac{a}{(1-r)}=10/(1-1/2)=10/(1/2)=20$$

Can you please explain to me how you got to the 10/(1-1/2) part in the last equation? I can not seem to trace the origin of the 1/2 part and why that expression is the divisor of 10.
Thank you.
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Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

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16 Jan 2017, 18:24
Given : A(n+1) = 10 + A(n)/2

A(2) = 10+ A(1)/2
A(3) = 10+ A(2)/2 = 15+A(1)/4
A(4) = 10+ A(3)/2 = 17.5+A(1)/8
A(5) = 10+ A(4)/2 = 18.75+A(1)/16
A(6) = 10+A(5)/2 = 19.375+A(1)/32
a(7) = 19.6875 + a(1)/64
so the second term for A(1000) somewhat 19.XxXXXXXx+A(1)/2^999 , the second part can be ignored closet answer would be 20.
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Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

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19 Jan 2017, 14:58
Given : A(n+1) = 10 + A(n)/2

A(2) = 10+ A(1)/2
A(3) = 10+ A(2)/2 = 15+A(1)/4
A(4) = 10+ A(3)/2 = 17.5+A(1)/8
A(5) = 10+ A(4)/2 = 18.75+A(1)/16
A(6) = 10+A(5)/2 = 19.375+A(1)/32
a(7) = 19.6875 + a(1)/64
so the second term for A(1000) somewhat 19.XxXXXXXx+A(1)/2^999 , the second part can be ignored closet answer would be 20.

I see now, throughout all equations you utilize A(1) and the end result is neglectable. Thanks a million!
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Re: There is a sequence A(n) where n is a positive integer such that A(n+1  [#permalink]

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27 Jul 2018, 01:54
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Re: There is a sequence A(n) where n is a positive integer such that A(n+1 &nbs [#permalink] 27 Jul 2018, 01:54
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