In the infinite sequence a1, a2, a3.....an, each term after

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In the infinite sequence a1, a2, a3.....an, each term after [#permalink]

04 Jan 2008, 05:07

In the infinite sequence a1, a2, a3.....an, each term after the first is equal to twice the previous term. if a5-a2=12, what is the value of a1?

a)4
b) 24/7
c) 2
d) 12/7
e) 6/7

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Re: Number Sequence [#permalink]

04 Jan 2008, 06:14

E

a1
a2=a1*2
a3=a1*2^2
...
an=a1*2^(n-1)
...

a5-a2=12 ==> a1*2^4-a1*2^1=12 ==> a1=12/(16-2)=6/7
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Re: Number Sequence [#permalink]

04 Jan 2008, 06:17

I even don't know what to say

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Re: Number Sequence [#permalink]

04 Jan 2008, 08:00

kazakhb wrote:

I even don't know what to say

Let's simplify this question:

The pattern says that each term is equal to twice previous term so:

Let A1 = X
Let A2 = 2x
Let A3 = 4x
Let A4 = 8x
Let A5 = 16x

or

A1 = X
A2 = 2x
A3 = 2^2(x)
A4 = 2^3(x)
A5 = 2^4(x)

(if this were a problem that involved much larger values, say A32 and A28 it would be beneficial to see this pattern as Ax = 2^(x-1)x as shown above. however, if we're dealing with A5 and A2 we can think of it the first way because it's the easiest approach)

Now that we have the values for A5 and A2 we can set up a formula and solve for x (which represents A1)

A5-A2 = 16x-2x = 12
14x = 12
x = 12/14 = 6/7

and BAM! You're finished and that's all there is to it!

Answer E

You'll have to forgive Walker. He's a certified math god and operates on a higher plane of quant logic than anyone I've seen :-D

. Walker can create an simply elegant formula for any problem you throw at him and solve it accurately.

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Re: Number Sequence [#permalink]

04 Jan 2008, 08:17

eschn3am wrote:

...and operates on a higher plane of quant logic than anyone....

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Re: Number Sequence [#permalink] 04 Jan 2008, 08:17

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In the infinite sequence a1, a2, a3.....an, each term after

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In the infinite sequence a1, a2, a3.....an, each term after

In the infinite sequence a1, a2, a3.....an, each term after

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