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# The sequence a1, a2, a3, ...an is such that a2 = a1, a3 =

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Manager
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The sequence a1, a2, a3, ...an is such that a2 = a1, a3 = [#permalink]

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26 Jan 2008, 00:08
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The sequence a1, a2, a3, ...an is such that a2 = a1, a3 = a1+a2,
an = a1 + a2 + a3......+a(n-1). If an = k, where n >2, a(n+3) =?

A) 2k
B) 4k
C) 6k
D) 8k
E) 16k
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26 Jan 2008, 01:00
D

$$a_n = a_1 + a_2 + a_3......+a_{n-1}$$

$$a_{n+1} = a_1 + a_2 + a_3......+a_{n-1}+a_n=a_n+a_n=2a_n$$

$$a_{n+m} =2^m*a_n$$

$$a_{n+3} =2^3*k=8k$$
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Senior Manager
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26 Jan 2008, 04:41
walker wrote:
D

$$a_n = a_1 + a_2 + a_3......+a_{n-1}$$

$$a_{n+1} = a_1 + a_2 + a_3......+a_{n-1}+a_n=a_n+a_n=2a_n$$

$$a_{n+m} =2^m*a_n$$

$$a_{n+3} =2^3*k=8k$$

D too.

We know that:
a2= a1
a3= a2+(a1)=a2+a2= 2a2
a4=a3+(a2+a1)=a3+a3=2a3

That is, we have that an=2an-1

Besides, we know that an=k
Then, an+3=2an+2=4an+1=8an=8k
Manager
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28 Jan 2008, 00:57
Yes. 8k is the answer. :D

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28 Jan 2008, 01:11
a1, a2, (a1+a2), 2*(a1+a2), 4*(a1+a2), 8*(a1+a2),...

For example: a3 = (a1+a2) = k

a(3+3)= a6 = 8*(a1+a2) = 8k

"D"
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28 Jan 2008, 03:23
GHIBI wrote:
The sequence a1, a2, a3, ...an is such that a2 = a1, a3 = a1+a2,
an = a1 + a2 + a3......+a(n-1). If an = k, where n >2, a(n+3) =?

A) 2k
B) 4k
C) 6k
D) 8k
E) 16k

D.

an = k
a(n+1) = 2k
a(n+2) = 2 x 2k = 4k
a(n+3) = 2 x 4k = 8k
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Re: PS - Sequence   [#permalink] 28 Jan 2008, 03:23
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