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For every integer k from 1 to 10, inclusive the
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01 Jan 2010, 18:46
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For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T is A. greater than 2 B. between 1 and 2 C. between 1/2 and 1 D. between 1/4 and 1/2 E. less than 1/4 i don't know how to solve it. it goes as (1/2)(1/2^2)+(1/2^3) ... (1/2^10)=T
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Re: Sequence: can anyone help with this question
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01 Jan 2010, 19:23
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T isA. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4 First of all we see that there is set of 10 numbers and every even term is negative. Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now. And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D. BUT there is shortcut:Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\). Now, the sum of infinite geometric progression with common ratio r<1[/m], is \(sum=\frac{b}{1r}\), where \(b\) is the first term.So, if the sequence were infinite then the sum would be: \(\frac{\frac{1}{2}}{1(\frac{1}{2})}=\frac{1}{3}\) This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\) (A, B and C are out). Also this means that the sum of our sequence is very close to \(\frac{1}{3}\) and for sure more than \(\frac{1}{4}\) (E out). So the answer is D.
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Re: Sequence: can anyone help with this question
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01 Jan 2010, 21:25
bekbek wrote: Thank you very much guys..
Any formula for this question:
For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T is
a) greater than 2 b) between 1 and 2 c) between 1/2 and 1 d) between 1/4 and 1/2 e) less than 1/4
OA is D but i don't know how to solve it. it goes as (1/2)(1/2^2)+(1/2^3) ... (1/2^10)=T Agree with Bunuel.. Use the Geometric Series formual to get the answer. S = a+ ar + ar^2+.....ar^n when 0<r<1 and series is infinte n=infinite a = a(1r^n)/(1r) = a/1r S=(1/2)(1/2^2)+(1/2^3) ... (1/2^10) = 1/2 (1+(1/2) +(1/2)^2 +(1/2)^3)....) = 1/2 ( 1/(1(1/2)) =1/3 So Values is near to 1/3 and slightly less than 1/3 D is the correct answer.
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Re: Sequence: can anyone help with this question
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02 Jan 2010, 02:39
the way i saw it is this way, the first two terms are 1/2 and 1/4, added together will give you 1/4. the rest of the term pairs will only be added to this, so there's no way that the end result will be less than 1/4. you now have your lowerbound...
> if you're pressed for time, you'll have a pretty good idea of the answer because D is the only one that has greater than 1/4, you can intelligently guess that... if you do have more time, press on...
if you look at the other term pairs, they'll be: 1/8, 1/16, 1/32 ... even if this goes on forever, the sequence will keep on halving itself...
now you do some quick calculations:
1/4 =2/8 = 4/16 = 8/32 .... etc. + 1/8 = 2/16 = 4/32 .... etc. + 1/16 = 2/32 .... etc. + 1/32 .... etc. _____________________________  add each column up = 1/4 3/8 7/16 15/32 .... etc.
what you'll notice is that the sums keep getting closer and closer to 1/2 but will never actually get there (if you graph this, it will be asymptotic to 1/2). you now have your upperbound... 1/2
answer: 1/4 < result < 1/2, D



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Re: Sequence: can anyone help with this question
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18 Feb 2011, 09:53



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Re: Sequence: can anyone help with this question
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18 Feb 2011, 12:09
A few things here: * I've never seen a real GMAT question that requires one to know any geometric sequence formulas. Of course there are questions where you might use such formulas, but there will always be a different approach available; * In any sequence question which gives an expression for each term, you'll always want to write down the first few terms using the given expression to work out what the sequence looks like. Here we have: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}, \frac{1}{128}, \frac{1}{256}, \frac{1}{512}, \frac{1}{1024}\) * Now, adding all of these fractions together would take a long time to do. The GMAT *never* requires you to perform any crazy calculations, so there must be a different way to answer the question. Notice the answer choices are only estimates, so we only need to estimate the sum of the first 10 terms. When we want to estimate the value of a sum, we ignore terms that make only a tiny contribution to the sum. The last few terms in our sequence are minuscule compared to the first few, so to get a good estimate, we can completely ignore them; adding, say, the first four terms will give a perfectly good approximation of the sum here (you get 5/16, which is enough to choose the right answer). * The sequence in this question is what is known as an 'alternating sequence'  that is, the terms alternate between positive and negative values. When adding an alternating sequence, you most often want to add your terms in pairs first, grouping one positive and one negative (add the 1st and 2nd term, the 3rd and 4th, and so on). One doesn't need to do this for this question, but it does make the answer a bit easier to see  we'd find our sum is \(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}\) from which you can instantly see the sum is greater than 1/4. Since every term here is tiny after the first two, the sum is certainly less than 1/2.
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Re: Sequence: can anyone help with this question
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Updated on: 25 May 2011, 09:54
Step 1: Lets substitute k=1 ((1)^2)*(1/(2^1)) = 1/2 Let substitute k=2 ((1)^3)*(1/(2^2)) = 1/4
Since we have understood the pattern, we can write the remaining numbers which are (1/21/4)+(1/81/16)+(1/321/64)+(1/1281/256)+(1/5121/1024)
Step 2: The initially value is 0.5 . From 0.5 we are subtracting 0.25. Then the remaining value is 0.25. Then we are adding 0.125 which gives 0.375 then we are subtracting 0.0625 which gives 0.3125. So, if we add further numbers also, the total will not cross 0.5 for the 10 numbers
So the clear option is (D)
Originally posted by mba4viplav on 25 May 2011, 05:40.
Last edited by mba4viplav on 25 May 2011, 09:54, edited 1 time in total.



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Re: Sequence: can anyone help with this question
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25 May 2011, 06:30
checking upto 5 terms will give the value.
(1/21/4)+(1/81/16)+(1/321/64)+ ....
1/4 + 1/16 + 1/64 + 1/128 + ..
value will be between 1/4 < x < 1/2.
D



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Re: Sequence: can anyone help with this question
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26 May 2011, 10:27
The first term is (1)^2*1/(2)^1 = 1/2 The second term is = 1/4 The third term is = 1/8 Looking at the answer choices, you don't need to continue. Since the denominator is increasing exponentially, the terms added and subtracted are becoming closer to 0. From the first term, we know we will never go above 1/2 After subtracting the second term, we know we will never go below ¼. Ans. D.
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Re: Sequence: can anyone help with this question
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27 Oct 2013, 22:41
Bunuel wrote: This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\)
So the answer is D.
To nitpick, this is actually incorrect. The series converges to 1/3 but not necessarily from the bottom. For example, a sum of 1 term (i.e. 1/2) is larger than 1/3, and so is sum of 3 terms. In fact it oscillates around 1/3. The real shortcut is to imagine a stretch from 0 to 1, and mark (mentally or with a pencil) where the sum with each additional term falls: 01 ____________________^ 1/2 01 ___________^ 1/2  1/4 = 1/4 01 ________________^ 1/4 + 1/8 = 3/8 01 _____________^ 3/8  1/16 = 5/16 You can clearly see that the first two terms established the boundaries at 1/2 and 1/4



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Re: Sequence: can anyone help with this question
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08 Jun 2014, 06:54
Bunuel wrote: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T is A. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4
First of all we see that there is set of 10 numbers and every even term is negative.
Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now.
And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D.
BUT there is shortcut:
Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\).
Now, the sum of infinite geometric progression with common ratio r<1[/m], is \(sum=\frac{b}{1r}\), where \(b\) is the first term.
So, if the sequence were infinite then the sum would be: \(\frac{\frac{1}{2}}{1(\frac{1}{2})}=\frac{1}{3}\)
This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\) (A, B and C are out). Also this means that the sum of our sequence is very close to \(\frac{1}{3}\) and for sure more than \(\frac{1}{4}\) (E out).
So the answer is D. Is there another way? Do we need to know the geometric series equation for the GMAT?



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Re: Sequence: can anyone help with this question
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08 Jun 2014, 10:05
ronr34 wrote: Bunuel wrote: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T is A. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4
First of all we see that there is set of 10 numbers and every even term is negative.
Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now.
And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D.
BUT there is shortcut:
Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\).
Now, the sum of infinite geometric progression with common ratio r<1[/m], is \(sum=\frac{b}{1r}\), where \(b\) is the first term.
So, if the sequence were infinite then the sum would be: \(\frac{\frac{1}{2}}{1(\frac{1}{2})}=\frac{1}{3}\)
This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\) (A, B and C are out). Also this means that the sum of our sequence is very close to \(\frac{1}{3}\) and for sure more than \(\frac{1}{4}\) (E out).
So the answer is D. Is there another way? Do we need to know the geometric series equation for the GMAT? 1. There are other ways given above. For example: foreveryintegerkfrom1to10inclusivethe88628.html#p8747432. Yes, there are certain problems for which knowing that formula would be helpful.
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For every integer k from 1 to 10, inclusive the
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02 Jul 2014, 18:17
IanStewart wrote: A few things here:
* I've never seen a real GMAT question that requires one to know any geometric sequence formulas. Of course there are questions where you might use such formulas, but there will always be a different approach available;
* In any sequence question which gives an expression for each term, you'll always want to write down the first few terms using the given expression to work out what the sequence looks like. Here we have:
\(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64}, \frac{1}{128}, \frac{1}{256}, \frac{1}{512}, \frac{1}{1024}\)
* Now, adding all of these fractions together would take a long time to do. The GMAT *never* requires you to perform any crazy calculations, so there must be a different way to answer the question. Notice the answer choices are only estimates, so we only need to estimate the sum of the first 10 terms. When we want to estimate the value of a sum, we ignore terms that make only a tiny contribution to the sum. The last few terms in our sequence are minuscule compared to the first few, so to get a good estimate, we can completely ignore them; adding, say, the first four terms will give a perfectly good approximation of the sum here (you get 5/16, which is enough to choose the right answer).
* The sequence in this question is what is known as an 'alternating sequence'  that is, the terms alternate between positive and negative values. When adding an alternating sequence, you most often want to add your terms in pairs first, grouping one positive and one negative (add the 1st and 2nd term, the 3rd and 4th, and so on). One doesn't need to do this for this question, but it does make the answer a bit easier to see  we'd find our sum is
\(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}\)
from which you can instantly see the sum is greater than 1/4. Since every term here is tiny after the first two, the sum is certainly less than 1/2. QUESTION : IT WAS A NEGATIVE (when we list out termsss) 1/4 why did you make it negative



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Re: For every integer k from 1 to 10, inclusive the
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14 Jun 2017, 15:18
bekbek wrote: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T is
A. greater than 2 B. between 1 and 2 C. between 1/2 and 1 D. between 1/4 and 1/2 E. less than 1/4 We are given that for every integer k from 1 to 10 inclusive, the kth term of a certain sequence is given by (1)^(k+1) x (1/2^k). We must determine the sum of the first 10 terms in the sequence. Before calculating the sum, we should recognize that the answer choices are provided as ranges of values, rather than as an exact value. Thus, we might not need to calculate the total of the 10 terms to determine an answer. Perhaps we can uncover a pattern to help us find the answer. Let’s start by listing out the first four terms. k = 1: (1)^(1+1) x (1/2^1) (1)^2 x 1/2 1 x 1/2 = 1/2 k = 2: (1)^(2+1) x (1/2^2) (1)^3 x 1/4 1 x 1/4 = 1/4 k = 3: (1)^(3+1) x (1/2^3) (1)^4 x 1/8 1 x 1/8 = 1/8 k = 4: (1)^(4+1) x (1/2^4) (1)^5 x 1/16 1 x 1/16 = 1/16 Recall that we are trying to estimate the value of T = 1/2 + (1/4) + 1/8 + (1/16) + … until we have 10 terms. In other words, T = 1/2 – 1/4 + 1/8 – 1/16 + … until there are 10 terms. We should notice that the absolute values of the terms are getting smaller: 1/2>1/4>1/8>1/16. Notice that starting from the first term of 1/2, we are subtracting something less than 1/2 (notice that 1/4 < 1/2), but then adding back something even less (notice 1/8 < 1/4), and the process continues. Thus, because 1/2 and 1/4 are the largest term and the smallest term in our set, respectively, the sum will never fall below 1/4 or exceed 1/2. Thus, we conclude that T is greater than 1/4 but less than 1/2. Answer: D
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Re: For every integer k from 1 to 10, inclusive the
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18 Aug 2018, 11:39
Bunuel wrote: BUT there is shortcut:
Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\).
Maybe a dumb question. How did you find the common ratio as 1/2?
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Re: For every integer k from 1 to 10, inclusive the
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21 Nov 2018, 02:22
Bunuel wrote: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T is A. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4
First of all we see that there is set of 10 numbers and every even term is negative.
Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now.
And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D. I am not able to understand that how this series in infinte, it's clearly given sume
BUT there is shortcut:
Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\).
Now, the sum of infinite geometric progression with common ratio r<1[/m], is \(sum=\frac{b}{1r}\), where \(b\) is the first term.
So, if the sequence were infinite then the sum would be: \(\frac{\frac{1}{2}}{1(\frac{1}{2})}=\frac{1}{3}\)
This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\) (A, B and C are out). Also this means that the sum of our sequence is very close to \(\frac{1}{3}\) and for sure more than \(\frac{1}{4}\) (E out).
So the answer is D. I am not able to understand that how this series is infinite, it's clearly given sum of integers from 1 to 10 inclusive? please clear my doubt



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Re: For every integer k from 1 to 10, inclusive the
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21 Nov 2018, 21:32
kunalkhanna wrote: Bunuel wrote: For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((1)^{(k+1)}*(\frac{1}{2^k})\) if T is the sum of the first 10 terms in the sequence, then T is A. Greater than 2 B. Between 1 and 2 C. Between 1/2 and 1 D. Between 1/4 and 1/2 E. Less than 1/4
First of all we see that there is set of 10 numbers and every even term is negative.
Second it's not hard to get this numbers: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... enough for calculations, we see pattern now.
And now the main part: adding them up is quite a job, after calculations you'll get \(\frac{341}{1024}\). You can add them up by pairs but it's also time consuming. Once we've done it we can conclude that it's more than \(\frac{1}{4}\) and less than \(\frac{1}{2}\), so answer is D. I am not able to understand that how this series in infinte, it's clearly given sume
BUT there is shortcut:
Sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)... represents geometric progression with first term \(\frac{1}{2}\) and the common ratio of \(\frac{1}{2}\).
Now, the sum of infinite geometric progression with common ratio r<1[/m], is \(sum=\frac{b}{1r}\), where \(b\) is the first term.
So, if the sequence were infinite then the sum would be: \(\frac{\frac{1}{2}}{1(\frac{1}{2})}=\frac{1}{3}\)
This means that no matter how many number (terms) we have their sum will never be more then \(\frac{1}{3}\) (A, B and C are out). Also this means that the sum of our sequence is very close to \(\frac{1}{3}\) and for sure more than \(\frac{1}{4}\) (E out).
So the answer is D. I am not able to understand that how this series is infinite, it's clearly given sum of integers from 1 to 10 inclusive? please clear my doubt "So, if the sequence were infinite then the sum would" summarises it. Even if you assume that its an infinite series you would get the same results
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