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PS- integers and number properties

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PS- integers and number properties [#permalink] New post 07 Jan 2006, 16:43
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Hey, here's another one from the gmat prep. I do not understand the question. Please help!

For every integer k from 1 to 10 inclusive, the kth term of a certain sequence is given by (-1)^(k+1) (1/2^k). If T is the sum of the first 10 terms, what is the value range for T?

note: -1 is raised to k+1, and the 2 of 1/2 is raised k

Thank you!
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 [#permalink] New post 07 Jan 2006, 17:18
This is an arithmetic/geometric series question.

k = 1 to 10
So,
1st term of the series = (-1)^2 x (1/2) = 1/2
2nd term of the series = (-1)^3 x (1/4) = -1/4
3rd term of the series = (-1)^4 x (1/8) = 1/8

Hence, common ratio r = -1/2 --> Geometric series
T is the sum of k terms and r < 1, so:
T = a[(1-r^k)/(1-r)] = 0.5[(1-(-0.5)^10)/91+0.5)] = 1023/(1024*3)
T ~= 1/3
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a lil confused [#permalink] New post 09 Jan 2006, 14:25
thanks!
the answer is "between 1/4 and 1/2"
so, how'd they get that?
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 [#permalink] New post 09 Jan 2006, 15:25
1/3 is between 1/4 and 1/2.
I vaguely remember this question and from the answer choices, I think the only one that accomodates 1/3 is "between 1/4 and 1/2".

hth :)
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 [#permalink] New post 09 Jan 2006, 19:20
Guys, the question asks for the value range for T.

This confused me at the start ..... value range = sum ??!?! I wouldve just calculated the first term and the tenth term, and then subtracted the two to find the 'value range' .... :?
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 [#permalink] New post 09 Jan 2006, 21:07
Its a Geometric progression with the common ration= -1/2

Sum = a(1-r^n) / (1-r)
Pluggin in values would give you 1/3
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 [#permalink] New post 10 Jan 2006, 09:12
How in the world do you guys figure this out? Can some one please explain this...

Hence, common ratio r = -1/2 --> Geometric series
T is the sum of k terms and r < 1, so:
T = a[(1-r^k)/(1-r)] = 0.5[(1-(-0.5)^10)/91+0.5)] = 1023/(1024*3)
T ~= 1/3


Thanks!
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 [#permalink] New post 10 Jan 2006, 17:49
Hi Chapman,

Have a look at this:
http://mathworld.wolfram.com/GeometricSeries.html

If the terms in a series have a common ratio, its a geometric series and there are generalised formulae for working out the nth term of the series and the sum of n terms.

If the terms have a common difference its an arithmetic series and again there are formulae for the nth term and sum of n terms.

So in the question above, I've worked out the common ratio (from the terms) and then used the formula for sum to get the value for T.

Hope that helps!
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 [#permalink] New post 11 Jan 2006, 06:20
Sorry, please bear with me. I haven't done math like this in 10 years.

Is "a" the absolute value of the ratio in T=a(1-r^n) / (1-r)?

Also, is "value range" another way of saying total or sum?

Thank you all very much for working through this!
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 [#permalink] New post 02 Mar 2006, 12:19
Chapman wrote:
How in the world do you guys figure this out? Can some one please explain this...

Hence, common ratio r = -1/2 --> Geometric series
T is the sum of k terms and r < 1, so:
T = a[(1-r^k)/(1-r)] = 0.5[(1-(-0.5)^10)/91+0.5)] = 1023/(1024*3)
T ~= 1/3


Thanks!

Can someone please explain the part highlighted.

where did 91 come from , should (1-r) = 1 - (-0.5)
I know I am missing an important point here.

Thanks


Thanks
  [#permalink] 02 Mar 2006, 12:19
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