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GMAT PREP sequence question help

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by kanha81 » Mon Mar 02, 2009 4:16 pm
k=1: first term = 1/2
k=2: second term = -1/4
...
k=10: tenth term = -1/2^10

T=summation of first 10 terms
= (1/2 - 1/4) + (1/8 - 1/16) + ... + (1/512 - 1/1024)

this is Geometric Progression

S(n) = a*(1-r^n) / (1-r)
here a=1/2; r=(-1/4) / (1/2) = -1/2; n=10

=> S(10) = (1/2) * (1-(-1/2)^10) / (1-(-1/2))
=> S(10) = (1/2) * (1-(1/2)) / (3/2)
= (1/2) * (1/2) / (3/2)
= 1/6 < 1/4

Hence [E]
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by semidevil » Mon Mar 02, 2009 6:58 pm
Is there a way to solve this w/out knowing that it's a geometric progression? I was adding it all by hand and it was taking forever. any shortcuts besides geometric progression?
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by joyseychow » Mon Mar 02, 2009 10:11 pm
semidevil wrote:Is there a way to solve this w/out knowing that it's a geometric progression? I was adding it all by hand and it was taking forever. any shortcuts besides geometric progression?
I did the same too. Can someone enlighten us on this?
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by kaf » Tue Mar 03, 2009 2:04 am
kanha81,

Thanks for the solution.

Stating that it was geometric progression was really helpful because all i had to do was to do a little research on geometric progression and found out what it was and as a result i got to know about arithmetric progression as well.

Thanks
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by Uri » Fri Mar 06, 2009 2:48 pm
we don't need G.P for this. once we simplify the function for different values ranging from 1 to 10 inclusive, we find a series as below:
(1/2 -1/4) + (1/8 - 1/16) + .....
Observe that the first group inside the bracket yields 1/4
the second group will yield 1/16
this will continue.
so, obviously the answer is more than 1/4. but it will never be 1/2, as we each successive group becomes even smaller than the previous one. So, ans will be (D).
hope this helps.
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