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# M17Q5

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Current Student
Joined: 27 Jun 2012
Posts: 418
Concentration: Strategy, Finance
Followers: 68

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18 Jan 2013, 17:36
The sum of infinite geometric progression with common ratio |r|<1 , is $$sum=\frac{b}{1-r}$$, where b is the first term.

Hence here $$sum = \frac{1}{2}/(1-\frac{1}{2}) = 1$$

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Re: M17Q5   [#permalink] 18 Jan 2013, 17:36

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# M17Q5

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