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GMAT prep question .

by rockeyb » Wed Nov 25, 2009 10:55 pm
For integer k from 1 to 10 inclusive , the k th term of certain sequence is given by [(-1)^(k+1)] * (1/2^k). If T is the sum of the first 10 terms in the sequence then T is :

a) greater than 2.
b) between 1 and 2 .
c) between 1/2 and 1 .
d) between 1/4 and 1/2 .
e) less than 1/4 .

Please explain how yo have arrived at the answer .

Thanks .
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by papgust » Wed Nov 25, 2009 11:45 pm
rockeyb wrote:For integer k from 1 to 10 inclusive , the k th term of certain sequence is given by [(-1)^(k+1)] * (1/2^k). If T is the sum of the first 10 terms in the sequence then T is :

a) greater than 2.
b) between 1 and 2 .
c) between 1/2 and 1 .
d) between 1/4 and 1/2 .
e) less than 1/4 .

Please explain how yo have arrived at the answer .

Thanks .

Its D IMO.

K=1 to 10 (inclusive). Plug in k=1,2,3.. until you find the pattern of the sequence.

k=1, (-1)^2 * 1/2^1 = 1/2
k=2, (-1)^3 * 1/2^2 = -1/4
..
..

From the sequence, we can notice that it is a Geometric Progression.

So, First term (a) = 1/2, ratio of difference (r) = -1/2, Number of terms (n) = 10

Sum of the terms of GP = (a * [1-r^n])/ 1-r {Since r < 1}

Substitute a and r in the formula,

1/2 * [1 - (-1/2)^10] / 1-(-1/2)

Solving this, you get 341/1024 which is approximately 0.33 or 1/3
which is between 1/2 and 1/4

Hence D

This is a traditional way of solving this problem. I can't think of any other method. Any short method for this is highly appreciated
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by rockeyb » Thu Nov 26, 2009 3:59 am
Hey Papgust ,

Thanks for your reply . I too did the same thing and took me about 8-10 mins to arrive this answer .I agree this is the traditional method and is a bit lengthy one too.

Guys is there and easy way to solve PLEASE HELP !!!!!!!!!!!!
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by rockeyb » Thu Nov 26, 2009 4:00 am
Hey Papgust ,

Thanks for your reply . I too did the same thing and took me about 8-10 mins to arrive this answer .I agree this is the traditional method and is a bit lengthy one too.

Guys is there any easy way to solve PLEASE HELP !!!!!!!!!!!!
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by Birgit Anne » Thu Nov 26, 2009 7:38 am
I think there is an easier way to do this:

Inserting the first three numbers for k we get:
k=1 --> 1/2
k=2 --> -1/4
k=3 --> 1/8
.
.
.

And so the numbers keep narrowing to 0. Therefore it is enough to add the first three numbers to know in what range the answer will be.

Hope this helps...
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