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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
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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 a certain
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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 a certain
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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 a certain
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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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Solution Kth Term.jpeg [ 24.64 KiB  Viewed 819 times ]
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
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26 Apr 2019, 00:54
Hi Well we all understood that series is \(\frac{1}{2}\) , \(\frac{1}{2^2}\) , \(\frac{1}{2^3}\), \(\frac{1}{2^4}\), \(\frac{1}{2^5}\) ,\(\frac{1}{2^6}\),\(\frac{1}{2^7}\) ,\(\frac{1}{2^8}\) ,\(\frac{1}{2^9}\) ,\(\frac{1}{2^10}\) Can i rewrite the series as {\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) }  { \(\frac{1}{2^2}\) + \(\frac{1}{2^4}\) + \(\frac{1}{2^6}\)+ \(\frac{1}{2^8}\) + \(\frac{1}{2^10}\) } {\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } \(\frac{1}{2}\) { \(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } Let {\(\frac{1}{2}\) +\(\frac{1}{2^3}\) +\(\frac{1}{2^5}\) +\(\frac{1}{2^7}\) +\(\frac{1}{2^9}\) } =q then \(q\frac{q}{2}\) \(\frac{q}{2}\) now q is in GP and we need to find the sum of GP whose first term is \(\frac{1}{2}\) and r is \(\frac{1}{4}\) , n is 5 Either you can use the formula for sum of infinte terms of GP or genereal formula . Lets try the general formula which says sum of n terms of GP we have \(\frac{\frac{1}{2} * (1  \frac{1}{4^5})}{ (1\frac{1}{4})}\) \(\frac{\frac{1}{2} * ( \frac{4^51}{4^5})}{ (\frac{3}{4})}\) Can we say \(\frac{4^51}{4^5}\) is nearly equal to 1 So this expression \(\frac{\frac{1}{2} * ( \frac{4^51}{4^5})}{ (\frac{3}{4})}\) hence simplifies to \(\frac{2}{3}\) now q= \(\frac{2}{3}\) we need the value of \(\frac{q}{2}\) = \(\frac{1}{3}\) So Choice D suits our Answer. PS: Does this solution look long  Yeah because i tried to explain each step. I am sure we will not being many steps that i wrote down . I did get my solution in 1: 03 min abhimahnaI guess you missed on signs in this post , lucky you ended getting the right answer. https://gmatclub.com/forum/kthtermof ... l#p1724824
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain
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10 Sep 2019, 20:19
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 Series has 10 terms where first term \(= \frac{1}{2}\) and 10th term \(=  \frac{1}{2^{10}}\). Every odd is positive and every even term is negative. The series becomes \(\frac{1}{2}  \frac{1}{4} + \frac{1}{8}  \frac{1}{16} + \frac{1}{32}  \frac{1}{64} + \frac{1}{128}  \frac{1}{256} + \frac{1}{512}  \frac{1}{1024}\) Separating positive terms and negative terms we have \(\frac{1}{2}  \frac{1}{4} + \frac{1}{8}  \frac{1}{16} + \frac{1}{32}  \frac{1}{64} + \frac{1}{128}  \frac{1}{256} + \frac{1}{512}  \frac{1}{1024}\) \(= \frac{1}{2}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)  \frac{1}{4}(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)\) \(= (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) * (\frac{1}{2}  \frac{1}{4})\) \(= \frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..)\) And \(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} .. > 1.25\) \(\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > \frac{1}{4} * 1.25\) \(\frac{1}{4} * (1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} ..) > 0.31\) Checking answers only D is correct since \(\frac{1}{4} < 0.31 < \frac{1}{2}\) Answer (D).
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Re: For every integer k from 1 to 10, inclusive, the kth term of
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07 Oct 2019, 00:58
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