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For integers n from 1 to 11, inclusive, the nth term of a ce
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14 Oct 2013, 08:37
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66% (01:34) correct 34% (03:56) wrong based on 213 sessions
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For integers n from 1 to 11, inclusive, the nth term of a certain sequence is given by Attachment:
Q.png [ 1.35 KiB  Viewed 4146 times ]
. 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.
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Re: For integers n from 1 to 11, inclusive, the nth term of a ce
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14 Oct 2013, 09:08
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 : \(\frac{1}{3}+\frac{1}{9} = \frac{13}{9} = \frac{2}{9} = 2*0.11(repeating) = 0.222222....\) 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 upto a value which is more than 0.25. D.
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Re: For integers n from 1 to 11, inclusive, the nth term of a ce
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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
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28 Oct 2013, 14:03
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/(1r) = 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
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05 Dec 2013, 10:01
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=\(\frac{a}{1r}\) , 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
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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
D



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Re: For integers n from 1 to 11, inclusive, the nth term of a ce
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06 Jan 2019, 18:09
BunuelIs the following a good approach: Step 1: Realize that there are 11 terms in total > means that 1^n will stay negative. This gets rid off A, B, and C. Step 2: n1 is (1/3) Step 3: n11 is  (1/(3^11)) > probably very close to 0. Dont know 3^11 so not even going to calculate. Step 4: other values will fall in between, so this means that D is the right choice




Re: For integers n from 1 to 11, inclusive, the nth term of a ce &nbs
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06 Jan 2019, 18:09






