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GMATPrep Test 1 #2_PS Number Properties #3

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by Anurag@Gurome » Sun Mar 11, 2012 8:26 pm
kwah wrote:What is the most efficient way to achieve the result for the question attached?

Answer: D

Thanks,
K

We know that the sum of n terms of a geometric series is given by:
S(n) = a(1 - r�)(1 - r), where a is the first term, r is the common ratio of the geometric progression and n = number of terms.

Here, a = 1/2, r = -1/2, n = 10
T = 1/2[1 - (-1/2)^10]/[1 + 1/2]
= 1/2[1 - 1/1024]/[3/2]
= 1/2 * 1023/1024 * (2/3)
= (1023/1024) * (1/3)
Now (1023/1024) = 1 approx, so T = 1/3, which lies between 1/4 and 1/2.

The correct answer is D.
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by dhiren8182 » Sun Mar 11, 2012 11:16 pm
Hi Anurag,
How did u get ratio and first term for the same??
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by Anurag@Gurome » Sun Mar 11, 2012 11:26 pm
dhiren8182 wrote:Hi Anurag,
How did u get ratio and first term for the same??
For k = 1, (-1)^k+1 * (1/2^k) = (-1)^(1+1) * (1/2^1) = 1/2
For k = 2, (-1)^k+1 * (1/2^k) = (-1)^(2+1) * (1/2^2) = -1/4
For k = 3, (-1)^k+1 * (1/2^k) = (-1)^(3+1) * (1/2^3) = 1/8
and so on...

So, first term = 1/2
Now, 2nd term/1st term = (-1/4)/(1/2) = -1/2 and 3rd term/2nd term = (1/8)/(-1/4) = -1/2
So, common ratio = -1/2

Hope that helps.
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