M17 # 05 : Retired Discussions [Locked]
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# M17 # 05

Author Message
Senior Manager
Joined: 29 Sep 2009
Posts: 396
GMAT 1: 690 Q47 V38
Followers: 2

Kudos [?]: 35 [0], given: 5

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03 Oct 2010, 12:31
What is $$\frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20}$$ between?

(C) 2008 GMAT Club - m17#5

* $$\frac{1}{2}$$ and $$\frac{2}{3}$$
* $$\frac{2}{3}$$ and $$\frac{3}{4}$$
* $$\frac{3}{4}$$ and $$\frac{9}{10}$$
* $$\frac{9}{10}$$ and $$\frac{10}{9}$$
* $$\frac{10}{9}$$ and $$\frac{3}{2}$$

This is a Geometric Progression:
First term(a) 1/2 , Common ratio(r) 1/2 and n : 20+1
Sum of a GP: a(1-r^n)/(1-r)
= 1/2 (1- (1/2)^21) / (1-1/2)
(1- (1/2)^21) should be close to 1
hence answer : 1/2(1)/(1/2) = 1 (close to 1)
D
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 956 [0], given: 25

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03 Oct 2010, 12:40
vicksikand wrote:
What is $$\frac{1}{2} + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + ... + \left(\frac{1}{2}\right)^{20}$$ between?

(C) 2008 GMAT Club - m17#5

* $$\frac{1}{2}$$ and $$\frac{2}{3}$$
* $$\frac{2}{3}$$ and $$\frac{3}{4}$$
* $$\frac{3}{4}$$ and $$\frac{9}{10}$$
* $$\frac{9}{10}$$ and $$\frac{10}{9}$$
* $$\frac{10}{9}$$ and $$\frac{3}{2}$$

This is a Geometric Progression:
First term(a) 1/2 , Common ratio(r) 1/2 and n : 20+1
Sum of a GP: a(1-r^n)/(1-r)
= 1/2 (1- (1/2)^21) / (1-1/2)
(1- (1/2)^21) should be close to 1
hence answer : 1/2(1)/(1/2) = 1 (close to 1)
D

Lower Bounds
First term is 1/2
If you add the first two terms, 1/2+1/4 you are above 3/4
So A & B can't be answers

Upper Bounds
If this was an infinite series, the sum would be (1/2)/(1-1/2) = 1
So this sum has to be less than 1
So E can't be the answer

Between C & D
You don't need to calculate the sum, just observe the pattern (half of the distance to 1 is always covered)
First term = 1/2
First two terms = 3/4
First three terms = 7/8
First four terms = 15/16
.. and so on ...

So it is easy to see that this sum will very quickly become greater than 9/10 (at the fourth term in fact)

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Posts: 547
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Kudos [?]: 61 [0], given: 562

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20 Feb 2013, 03:44
Bunuel,
Is there any way to solve this without using the formula?

Also, is GP tested on GMAT?
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Kudos [?]: 93051 [0], given: 10541

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26 Feb 2013, 07:19
Sachin9 wrote:
Bunuel,
Is there any way to solve this without using the formula?

Also, is GP tested on GMAT?

Check here: m17q5-72268.html
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Re: M17 # 05   [#permalink] 26 Feb 2013, 07:19
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# M17 # 05

Moderator: Bunuel

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