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Intern
Joined: 04 Sep 2009
Posts: 43
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Kudos [?]: 54 [0], given: 9

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28 May 2011, 16:02
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In a series of 10 numbers the addition of all 10 numbers is 3069 and each number is two times the previous one. What is the value of the 3rd number?

Just expecting to figure out another way to do it... maybe faster

[Reveal] Spoiler:
ans: 12

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Joined: 24 Mar 2011
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Kudos [?]: 167 [1] , given: 20

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28 May 2011, 17:24
1
KUDOS
we know the series, its 2^0x, 2^1x, 2^2x, .... 2^10x

i think the other way to solve this is -

geometric progression, sum of 10 numbers = a (1-r^n)/1-r
3069 = 2^0 x (1-2^10)/1-2
solving it for x, x = 3

2^2x = 12
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Joined: 16 Nov 2010
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29 May 2011, 01:35
Sum of a GP = a(r^n - 1)/(r - 1)

r = 2, n = 10

3069 = a(2^10 - 1)/(2-1)

=> a = 3069/1023 = 3

=> 3rd term = 3 * 2^2 = 12
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Joined: 08 Jan 2009
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GMAT 1: 770 Q50 V46
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Kudos [?]: 136 [1] , given: 7

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29 May 2011, 01:43
1
KUDOS
I think when you start writing out the terms the pattern becomes fairly apparent, and you can sum mentally without much difficulty by treating the first term as a two, and then remembering to minus one.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512

Treat as:

2 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512
4 + 4 + ...
8 + 8 + ...
512 + 512 = 1024

Adjust for the one, 1024 - 1 = 1023.

x = 3069 / 1023 = 3

4x = 12
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Joined: 12 Oct 2009
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29 May 2011, 12:11
Using GP formulae for Sum and Nth term - 12
Re: Series   [#permalink] 29 May 2011, 12:11
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