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# Numbers in Sequence

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Manager
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Numbers in Sequence [#permalink]

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04 Aug 2009, 20:20
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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?

4
2
12/7
6/7

I keep getting the value 1 as the answer to the question. I am not sure what I am doing wrong.

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Intern
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Re: Numbers in Sequence [#permalink]

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04 Aug 2009, 23:21
Lets assume the first number is x. The series would then be x,2x,4x,8x,16x, and so on.

The question gives the equation as A5-A2=12

Replace valuse of x in this:

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

This is the correct answer
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Intern
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Re: Numbers in Sequence [#permalink]

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05 Aug 2009, 01:18
Another method could be to use the geometric series (GS) formula.

Nth term in GR = ar^(n-1), where a is the first term in the series (a=A1), r is the rate of increase and n is the number of the term.

As A5 - A2 = 12,
the equation becomes:
a(2)^(5-1) - a(2)^(2-1) = 12
a(2)^4 - a(2)^1 = 12
a(2) (2^3 - 1) = 12
a(2) 7 = 12
a=12/14=6/7

Therefore, A1=6/7
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Re: Numbers in Sequence [#permalink]

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05 Aug 2009, 05:23
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?

4
2
12/7
6/7

A5 = 2A4 = 4A3=8A2

8A2-A2 = 12 , 7A2 = 12 , A2 = 12/7 = 1 5/7

A2 = 2A1 thus A1 = 12/7*1/2 = 6/7

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Re: Numbers in Sequence [#permalink]

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05 Aug 2009, 13:07
aye, 6/7 is the answer....A1*(2^4 - 2^1) = 12, A = 12/14 = 6/7

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Manager
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Re: Numbers in Sequence [#permalink]

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05 Aug 2009, 21:11
the series will be a1, 2a1, 4a1, 8a1, 16a1...given that 16a1 - 2a1 = 12 hence a1 = 6/7..OA D
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Re: Numbers in Sequence   [#permalink] 05 Aug 2009, 21:11
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# Numbers in Sequence

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