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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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15 Dec 2015, 12:04
prathns wrote: For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1)*(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 1. Calculate fist let's say 4 terms to see the pattern: so the max value is 1/2 (the 1st value) the 4th value is 11/32 which is < 1/2 2. Look at the answer choices: from 1 we know that max value is 1/2 and the last value will be less than 1/2 > the only answer choice, in which 1/2 is max value is (D) Answer D
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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02 Aug 2016, 20:32
Quick shortcut to find Sum after finding out the initial terms and the fact that its in GP (common ratio is \(\frac{1}{2}\)) \(S = \frac{1}{2}  \frac{1}{4} + \frac{1}{8}  \frac{1}{16} ...\) Multiply both sides by the common ratio \(\frac{1}{2}\): \((\frac{1}{2})*S =  \frac{1}{4} + \frac{1}{8}  \frac{1}{16} + \frac{1}{32} ...\) \((\frac{1}{2})*S = S  (\frac{1}{2})\) \(S = \frac{1}{3}\)



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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25 Aug 2016, 14:05
One of those Amazing GMAT prep Questions here is what i did => HEre the series is => 1/21/4+1/161/32+1/64 and so on Here rewriting the terms => 1/4+1/41/4+1/32+1/321/32+1/64+1/641/64.... => 1/4+1/32+1/64+1/128+1/256 The sum of a GP with r<1 => a(1r^n)/1r => 1/4*(1(1/4)^5)/3/4 => 1/3 1/(4^5 *3) => 0.33(0.00whocares) => 0.30 approx Hence the sum has to be >1/4 but less than 1/2 => Smash that D Alternatively sum of an infinite GP series => a/1r => 1/4/3/4 => 1/3 => 0.33333333333 => SWEET. Smash that D again
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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24 Dec 2016, 12:30
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 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 \(r<1\), 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. Answer: D. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661 Amazing solution Bunuel.
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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15 Jan 2017, 11:48
Seriously  would anyone be able to resolve in 2 minutes?



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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15 Jan 2017, 21:58
Hi Erjan_S, If you were trying to calculate the exact sum of this sequence, then you would likely find it almost impossible to do that in under 2 minutes. Thankfully, this question doesn't actually ask you to do that  the answer choices are all RANGES, which is a big 'hint' that you're supposed to something OTHER than calculate the exact sum. The 'key' to this question is to look at the sequence in 'pairs' (re: the 1st and 2nd, the 3rd and 4th, the 5th and 6th, etc.). Defining how pairs of terms relate to one another makes solving this question a lot easier than trying to calculate the sum of all 10 terms (my solution explains all of this in detail). GMAT assassins aren't born, they're made, Rich
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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22 Feb 2017, 21:39
Hi Bunuel, Can we say that if r <1, then it's an infinite GP ? How does one define infinite GP? When does one use the forumula Sum = b1 (r^n1)/r1 ? Please explain.. Bunuel wrote: Stiv wrote: \(\frac{first \ term}{1constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)? The sum of infinite geometric progression with common ratio \(r<1\), is \(sum=\frac{b}{1r}\), where \(b\) is the first term.



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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23 Feb 2017, 00:15



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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23 Feb 2017, 01:16
cuhmoon wrote: Hi Bunuel, Can we say that if r <1, then it's an infinite GP ? How does one define infinite GP? When does one use the forumula Sum = b1 (r^n1)/r1 ? Please explain.. Bunuel wrote: Stiv wrote: \(\frac{first \ term}{1constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)? The sum of infinite geometric progression with common ratio \(r<1\), is \(sum=\frac{b}{1r}\), where \(b\) is the first term. To add to what Bunuel said, r is the common ratio and has nothing to do with whether a progression is infinite or finite. In either case, r can be less than 1 or more than 1 e.g. 1, 3, 9, 27, 81 ... (infinite sequence with r = 3) 27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3) 1, 3, 9 (finite sequence with 3 elements and r = 3) 27, 9, 3 (finite sequence with 3 elements and r = 1/3) Now the point is that the sum of all terms of the first sequence is infinite. The terms will can getting larger and will keep adding up. So the sum will be infinite. We can find the exact sum of the rest of the 3 sequences. 27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3) Sum = a/(1  r) = 27/(1  1/3) = 81/2 1, 3, 9 (finite sequence with 3 elements and r = 3) Sum = a(r^n  1)/(r  1) = 1*(3^3  1)/(3  1) = 13 27, 9, 3 (finite sequence with 3 elements and r = 1/3) Sum = a(1  r^n)/(1  r) = 27*(1  1/3^3)/(1  1/3) = 26*3/2 = 39
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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28 Mar 2017, 09:31
let's try to plug in some numbers in the beginning: we see that 1*1/2=50% 1*1/4=25% 1*1/8=13% 1*1/16=7% 1*1/32=3,5% 1*1/64=2,25% By adding them up we will get somewhere at around 33%. Hence 1/3 The answer is D then



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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04 Apr 2017, 12:58
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 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 \(r<1\), 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. Answer: D. Other solutions at: http://gmatclub.com/forum/sequencecan ... ml#p668661 Hi Bunuel, Thank you for the explanation. Can you please elaborate on the last part. Since we have found the sum of an infinite series how can we say that the sum of just 10 terms will be very close to (1/3) and for sure more than (1/4). I understand everything else but just finding this part difficult to rule out. Can you please help! Regards, Shradha



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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15 May 2017, 19:36
TehJay wrote: prathns wrote: For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1) *(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 have no clue what info has been given and how to use it to derive T.
Kindly post a detailed explanation.
Thanks. Prath. When doing sequence problems, it usually helps to look at at least the first few terms. So in this case: \(a_k = (1)^{k+1} * \frac{1}{2^k}\) This gives us: \(a_1 = \frac{1}{2}\) \(a_2 = \frac{1}{4}\) \(a_3 = \frac{1}{8}\) \(a_4 = \frac{1}{16}\) We can stop there. The first, and largest, term is 1/2, and we then subtract 1/4. We will then add and subtract fractions that will continue to get smaller and smaller. So we can immediately eliminate A, B, and C, since the sum cannot possibly be greater than 1/2. Now we're left with D and E. Note that the sum of the first two terms is 1/4. Then you add another 1/8 and subtract 1/16. This pattern will continue all the way through the tenth term, and you should be able to see that there's no way this sum will become less than 1/4. So the answer is D. It's clear the sum will be above 1/4; however, if the progression was continued, wouldn't the sum eventually exceed 1/2, since we are continually adding positive fractions every 2 progressions.



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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15 Jun 2017, 22:26
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 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 \(r<1\), 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. Answer: D. Other solutions at: http://gmatclub.com/forum/sequencecan ... ml#p668661 Bunuel, how come the ratio here is 1/2, and not 1/2? If there are alternating signs in a sequence, is the "ratio" always the negative value? thanks



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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15 Jun 2017, 22:55
iyera211 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 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 \(r<1\), 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. Answer: D. Other solutions at: http://gmatclub.com/forum/sequencecan ... ml#p668661 Bunuel, how come the ratio here is 1/2, and not 1/2? If there are alternating signs in a sequence, is the "ratio" always the negative value? thanks Common ratio is a ratio of consecutive terms. Divide two consecutive terms what do you get? 1/2 or 1/2?
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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16 Jun 2017, 07:05
prathns wrote: For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1)*(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 List some terms to see the pattern. We get: T = 1/2  1/4 + 1/8  1/16 + . . . Notice that we can rewrite this as T = (1/2  1/4) + (1/8  1/16) + . . . When you start simplifying each part in brackets, you'll see a pattern emerge. We get... T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024Notice that 1/64, 1/256, and 1/1024 are each less than 1/16 So, ( 1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16) Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = ( a number less than 1/4) Now start from the beginning: T = 1/4 + ( 1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + ( a number less 1/4) = A number less than 1/2 Of course, we can also see that T > 1/4 So, 1/4 < T < 1/2 Answer: Cheers, Brent
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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25 Jun 2017, 11:06
prathns wrote: For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (1)^(k+1)*(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 Refer to the solution in the picture
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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06 Sep 2017, 07:09
Bunuel wrote: Stiv wrote: \(\frac{first \ term}{1constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)? The sum of infinite geometric progression with common ratio \(r<1\), is \(sum=\frac{b}{1r}\), where \(b\) is the first term. Hi Bunuel, Can you please direct me to a link which describes the recent trend(type) of questions coming in the GMAT in the recent months? I mean to say, any stress on particular chapters and obsolete chapters if any.



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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28 Feb 2018, 07:45
I accidentally found a very elegant solution:)
Series is T = 1/2  1/2^2 + 1/2^3  1/2^4 + 1/2^5  1/2^6 +..
This series can be written in two ways:
i) T = 1/2  (1/2^2  1/2^3)  (1/2^4  1/2^5)  (1/2^6 +..)... T= 1/2  1/2^2  1/2^5  1/2^7..
So, we clearly know that T is less than 1/2
ii) T = 1/2  1/2^2 + 1/2^3  1/2^4 + 1/2^5  1/2^6 +.. T = (1/2  1/2^2) + (1/2^3  1/2^4) + (1/2^5  1/2^6) +.. T = 1/2^2 + 1/2^4 + 1/2^6+.. T = 1/4 + 1/2^4 + 1/2^6+..
So, we clearly know that T is more than 1/4
Hence, from i) and ii), it is clear that I is between 1/4 and 1/2.




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