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For integers n from 1 to 11, inclusive, the nth term of a ce

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For integers n from 1 to 11, inclusive, the nth term of a ce [#permalink]

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New post 14 Oct 2013, 08:37
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For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by
Attachment:
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.
If S is the sum of the first 11 terms in the sequence then S is

(A) Greater than 2

(B) Between 1 and 2

(C) Between 0 and 1/2

(D) Between -1/2 and 0

(E) Less than -1/2

I got the answer correct but i am looking for the easiest and fastest method to solve this.
[Reveal] Spoiler: OA
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Re: For integers n from 1 to 11, inclusive, the nth term of a ce [#permalink]

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New post 14 Oct 2013, 09:08
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praffulpatel wrote:
For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by
Attachment:
Q.png
.
If S is the sum of the first 11 terms in the sequence then S is

(A) Greater than 2

(B) Between 1 and 2

(C) Between 0 and 1/2

(D) Between -1/2 and 0

(E) Less than -1/2

I got the answer correct but i am looking for the easiest and fastest method to solve this.


Method I :Crude

The fastest way is to realize that

1. The sign of the term with the greatest absolute value( that is the first term) has a negative sign attached to it
2. The given range is all within the precision of 1 decimal value,i.e. between 0.0 and -0.5 or less than -0.5
3.The greater the denominator, the more will be the no of decimal places

Take the sum of the first two terms : \(-\frac{1}{3}+\frac{1}{9} = \frac{1-3}{9} = \frac{-2}{9} = -2*0.11(repeating) = -0.222222....\)
Rest of the remaining terms, can ONLY change the above value after the first decimal place. Thus, the value would hover around -0.2XXXX and thus the answer is D.

Method II :

A little observation and fair pairing of the terms would show that the sum of the first two terms : \(\frac{-2}{3^2}\) , the pair of the 3rd and 4th terms : \(\frac{-2}{3^4}\) and so on.

Take an infinite sum of the series with terms as \(\frac{-2}{3^2} , \frac{-2}{3^4},.. \to -2*(\frac{1}{3^2}+\frac{1}{3^4}+...) = -2*\frac{\frac{1}{3^2}}{1-\frac{1}{9}\) = \(-2*\frac{1}{8} = -0.25\). Thus, the given series can never sum up-to a value which is more than -0.25.

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Re: For integers n from 1 to 11, inclusive, the nth term of a ce [#permalink]

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New post 26 Oct 2013, 06:24
praffulpatel wrote:
For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by
Attachment:
Q.png
.
If S is the sum of the first 11 terms in the sequence then S is

(A) Greater than 2

(B) Between 1 and 2

(C) Between 0 and 1/2

(D) Between -1/2 and 0

(E) Less than -1/2

I got the answer correct but i am looking for the easiest and fastest method to solve this.


We have \(S = \sum_{n=1}^{11} \frac{(-1)^n}{3^n} =\frac{-1}{3} \sum_{n=0}^{10} \frac{(-1)^n}{3^n} = \frac{-1}{3} \frac{ 1 - \frac{(-1)^{11}}{3^{11}}}{1 - \frac{-1}{3}}\). So S is approximately equal to \(\frac{-1}{3} \frac{ 1 }{\frac{4}{3}} = \frac{-1}{4}\).
Therefore, the answer is D
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Re: For integers n from 1 to 11, inclusive, the nth term of a ce [#permalink]

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New post 28 Oct 2013, 14:03
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The series is -1/3+1/3^2...So the value of this series has to be higher than -1/3
Further, this is a GP with r=-1/3
Assuming this goes on till infinity the sum would equal a/(1-r) = -1/3/(1-(-1/3))=-1/4
hence the sum of this series has to lie between -1/3 and -1/4
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Re: For integers n from 1 to 11, inclusive, the nth term of a ce [#permalink]

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New post 05 Dec 2013, 10:01
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The sequence is given by (-1)^n*(1/3^n)
Putting n=1, the above sequence gives us the value, -1/3
Putting n=2, the above sequence gives us the value 1/3^2
Putting n=3, the above sequence gives us the value -1/3^3
So it forms a sequence -1/3,1/3^2,-1/3^3,1/3^4....
The above sequence will turn out to be in Geometric Progression(GP) with common ration (r) = -1/3

The easiest way to solve is by following the below approach which says that.
If the absolute value of common ratio |r|<1, then the Sum of GP is defined by
Sum=\(\frac{a}{1-r}\) , where a= First term of the GP, and r = common ratio


As we can see the 'r' in our sequence is -1/3, hence |r|<1, so the sum of GP can be given as
\(S =\frac{-1}{3}/1-(\frac{-1}{3}) => \frac{-1}{4}\)

Now look out the answer choices.
Answer D is the correct answer since -1/4 is between -1/2 and 0
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For integers n from 1 to 11, inclusive, the nth term of a ce [#permalink]

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New post 06 Sep 2015, 20:56
First term=-1^1*(1/3)=-1/3
Second term=-1^2*(1/9)=1/9,
Sum=-1/3+1/9=-2/9, so every pair (3, 4th; 5,6th) sum will be negative. Total sum will be negative and in -1/3<S<0

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Re: For integers n from 1 to 11, inclusive, the nth term of a ce [#permalink]

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New post 05 May 2017, 15:12
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Re: For integers n from 1 to 11, inclusive, the nth term of a ce   [#permalink] 05 May 2017, 15:12
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