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26 Apr 2012, 08:07



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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27 Oct 2012, 07:44
simple explanation: look at the first term and second term: it says quarter is removed from half, result would be less than half of first term.look at the next two terms: already the terms are less than half and quarter, result would much more less than quarter ( half of the first term in the sequence terms). similarity, adding all the results, no mater how many terms are there, ultimate result would be always less than 1/2 and more than 1/4. but, if there are more terms it may happen that result would much more more than small term. example more than small term 1/16.



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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17 Nov 2012, 03:05
gmat dose not test the memory of formular. HOw to solve 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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18 Nov 2012, 11:15
thangvietnam wrote: gmat dose not test the memory of formular.
HOw to solve in 2 minutes. ? Check out the explanation of this question here: http://www.veritasprep.com/blog/2012/03 ... sequences/
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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 2012, 00:32
some really good explanations for this question! I was wondering do we need to know the infinite sequence formula  you can solve the question in a minute if you know the formula. The graphic approach was really good too...
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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 2012, 00:39



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Re: Numbers , Squences , Indices [#permalink]
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23 Apr 2013, 00:47
First term in the series is 1/2, second 1/4 third 1/8 fourth is 1/16. and likewise till 10th term.
so sum of the series is = 1/21/4+1/81/16....
so it is essentially the sum of differences between two adjcent terms((1/21/4)+(1/81/16) till 10th term.
so it will be 1/4 + 1/16 + etc.. we can stop calculating after this. 1/4 = .25 and 1/16 = .0625 and all the subsequent terms will be much less than .0625.
so the sum will be .25+.0625+etc.. = .3abcd
hence its between 1/4 and 1/2
it took me around a minute to solve it. its a sub 1.5 minute question i believe.



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Re: Numbers , Squences , Indices [#permalink]
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23 Apr 2013, 02:01
kabilank87 wrote: For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?
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
\(S = (\frac{1}{2}+\frac{1}{2^{3}}+..\frac{1}{2^{9}})  (\frac{1}{2^{2}}+..\frac{1}{2^{10}})\) Multiply on both sides by 2: \(2S = 1+ (\frac{1}{2^{2}}+..\frac{1}{2^{8}})(\frac{1}{2}+\frac{1}{2^{3}}+..\frac{1}{2^{9}})\)Add both the equations: \(3S = 1\frac{1}{2^{10}}\) \(S = \frac{1}{3}\frac{1}{3*2^{10}}\) D.
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Re: Numbers , Squences , Indices [#permalink]
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23 Apr 2013, 03:57
Zarrolou wrote: kabilank87 wrote: For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?
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 answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ? Here we need to find a pattern \(\frac{1}{2},\frac{1}{4},\frac{1}{8},...\) as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms. We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us 1/4, the second pair is 1/81/16 positive so we add value to 1/4, so the sum will be greater.(this is true also for the next pairs, so we add to 1/ 4 a positive value for each pair) D Hope its clear, let me know Hi Zarro , Very clear. But how to approach this kind of problems in GMAT without taking much time. Even though you explanation look very simple and time saving , i am not sure how i respond to the question same way as you explained. Do you have any siggestions how can i approach these problems ? Thanks in advance.
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Re: Numbers , Squences , Indices [#permalink]
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23 Apr 2013, 04:11
A quick way to do this : the sum of the sequence for k from 1 to 10 is : \(Sum = 1/2  1/(2^2) + 1/(2^3) ....\) Notice that ... represents a very small numbers that even substracted or added to the first three terms (1/2 , 1/4 and 1/8), we can neglect it . Hence, since 1/21/4+1/8 is between 1/4 and 1/2 , the sum will be between 1/4 and 1/2 as well. Answer : D
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Re: Numbers , Squences , Indices [#permalink]
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23 Apr 2013, 04:13
kabilank87 wrote: Hi Zarro ,
Very clear. But how to approach this kind of problems in GMAT without taking much time. Even though you explanation look very simple and time saving , i am not sure how i respond to the question same way as you explained. Do you have any siggestions how can i approach these problems ?
Thanks in advance.
Hi kabilank87, when we deal with a series (as in this case) the first and most important thing to do is find a pattern. One you've found that you can continue the series with no limit ( the GMAT will never ask you the exact value of a seires such this one), but the role of patterns is crucial also in question where you're asked to find the \(N^t^h\) term of a sequence.
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Re: Numbers , Squences , Indices [#permalink]
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24 Apr 2013, 12:21
Zarrolou wrote: kabilank87 wrote: For every integers K from 1 to 10 inclusive, the K th term of a certail sequence is given by [(1)^(K+1)](1 / (2^K). What is the sum of first 10 terms of the sequence ?
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 answer this question in my second attempt, but it takes a lot of times around 4 minutes. Please explain a shorter way to do this ? Here we need to find a pattern \(\frac{1}{2},\frac{1}{4},\frac{1}{8},...\) as you see the sign changes every term. The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms.We can eliminate any option that gives us a upper limit greater than 1/2. We are down to D and E. Is the sum less than 1/4? Take the sum of pair of terms : the first 2 give us \(\frac{1}{4}\), the second pair is \(\frac{1}{8}\frac{1}{16}\) positive so we add value to \(\frac{1}{4}\), so the sum will be greater.(this is true also for the next pairs, so we add to \(\frac{1}{4}\) a positive value for each pair) D Hope its clear, let me know Can you please elaborate more , Zarrou ? I still do not understand , and the line in red tricks me a lot . Thanks in advance
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Re: Numbers , Squences , Indices [#permalink]
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24 Apr 2013, 12:47
Zarrolou wrote: TheNona wrote: Can you please elaborate more , Zarrou ? I still do not understand , and the line in red tricks me a lot . Thanks in advance The pattern: \(\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},...\) We have to sum those elements so: \(\frac{1}{2}\frac{1}{4}+\frac{1}{8}\frac{1}{16}+...\) The first term is \(\frac{1}{2}\), to this we subtract 1/4, to the result we add 1/8, and so on As you see the operations involve smaller and smaller term each time. The first thing to notice here is that the sum will be <1/2, we can easily see this: \(\frac{1}{2}\frac{1}{4}=\frac{1}{4}\) and the operations will not produce a result >1/2. Hope it's clear here: the numbers decrease too rapidly to produce a result as big as the first term! Now we are left with D and E: the only 2 option which result is <1/2. And the question is: will the sum be less than 1/4? We have to find an easy way to see this, consider this fact: \(\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},...\) take the sum of couple of terms: 1st with 2nd, 3rd with 4th, and so on... The result will be positive for each couple, lets take a look:\(\frac{1}{2}\frac{1}{4}=\frac{1}{4}\) for the first one, \(+\frac{1}{8}\frac{1}{16}=\frac{1}{16}(>0)\) and so on. The thing to take away here is: 1/4+(num>0)+(num>0)+... will NOT be less than 1/4, how could it be if all numbers are positive? So the sum will be GREATER than 1/4 and LESSER than 1/4. Hope everything is clear now, I have been as exhaustive as possible, let me know Perfect! Thanks a lot
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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11 Oct 2013, 13:22
Since I solved it from a different method than mentioned here, thought it to share.
I guess we all might have deudced there are 10 terms and alternately positive and negative.
I tried with GP sum formula and got lost in calculation.
Since we have alternately +  we can make use of it.
Take 1st term , 2nd term 1/2 and 1/4, add them to get 1/4 Similarly 3rd and 4rth term gives you 1/8 ( 1/8 + 1/16) 1/16 We see a multiplication pattern of 4 here so no need to calculate further.
1/4, 1/16, 1/64, 1/256, 1/1024
Add them to get 256+64+16+4+1= 341/1024
Clealry less than half so lies b/w 1/4 and 1/2
Not a very great method but I guess helps me avoid calculcation mistake if I go for GP sum.



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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 2014, 23:18
Question is written in such a way that it is difficult to comprehend but once it is then it is just a matter of few seconds to crack.Here my answer.
T(K) = (−1)^k+1 * 1/2^k
So, T(1) =1/2 T(2)= 1/4 T(3)= 1/8 T(4)= 1/16 .... and so on No need to calculate higher terms because they doesn't produce any significant increase in sum as answers are widely distributed.
Sum=(1/2) + (1/4) +(1/8) +(1/16)+..... =>1/4 +1/16 => 0.25+0.0625 => .3125
SO answer is between 0.25 and 0.5 . option D
Option A,B & E can be rejected just by looking terms as they are too big or small enough.
Proper calculation approach: even though u calculate each term Sum=1/4+1/16+1/64+1/256+1/1024 =>0.25+0.0625+0.015625+0.00390625 +0.000976 =0.3330



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Exponents / powers (2) [#permalink]
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13 Mar 2014, 08:17
How can I approach this question?
(1)k+1(½k). T is the sum of the first 10 k, is t a. > 2 b. between 1 and 2 c. between ½ and 1 d. between ¼ and ½ e. < ¼



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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12 Apr 2014, 00:02
Sequence will be 1/21/4+1/81/16.......
Sum of first pair= 1/4 = 25% Sum of 2nd pair=1/16 = Appropriately 6%
So we can see that % is reducing. Therefore next three percentage each of which will not be more than 6%
So, Sum= 25%+6%+ ........ = in between (25 and 50)% D shows that in percentage.



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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05 Nov 2014, 22:58
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[/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. Answer: D. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661 Alternative, if you use the geometric series formula. S = \frac{a(1r^n)}{1r} where a = first term, r = multiple factor, n = # of terms. Hi Bunuel, how are these two formula different? Thank you.



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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]
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06 Nov 2014, 06:46
vietnammba 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[/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. Answer: D. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661 Alternative, if you use the geometric series formula. S = \frac{a(1r^n)}{1r} where a = first term, r = multiple factor, n = # of terms. Hi Bunuel, how are these two formula different? Thank you. The formula I used is for the sum of infinite geometric progression with common ratio r<1.
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