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For every integer k from 1 to 10, inclusive, the kth term of
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07 Jan 2010, 06:22
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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
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For every integer k from 1 to 10, inclusive, the kth term of
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07 Jan 2010, 07:30
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
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Re: Sequence
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07 Jan 2010, 09:13
Alternative, if you use the geometric series formula.
S = \(\frac{a(1r^n)}{1r}\)
where a = first term, r = multiple factor, n = # of terms.
So in this case, a = 1/2, r = 1/2, n = 10.
Therefore,
S=( (1/2) * (1 (1/2)^10) ) / (1(1/2) S=( (1/2) * (1  (1/1024) ) / (3/2) S=(11/1024) / 3 S=(1023/1024 ) / 3 Since 1023/1024 is close to 1, dividing it by 3 would get us to approximately 1/3, which is between 1/2 and 1/4. So the answer is D.




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Re: Sequence
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27 Apr 2010, 19:08
Nice question and nice explanations! I was more leaning toward geometric series solution.
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Re: DS : Number Theory
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10 Jul 2010, 01:10
Kth term: A(k) = (1) ^ (k+1) * 1 / 2^k
T is the sum of the first 10 terms of this sequence.
A(1) = (1)^(1+1)*1/2^1 = 1/2
A(2) = (1)^(1+2)*1/2^2 = 1/4
You can see that for odd k it would be positive values of \(1/2^k\) and for even k it would be negative values of \(1/2^k\)
So it's like this:
\(T = \frac{1}{2}  \frac{1}{4} + \frac{1}{18}  \frac{1}{16}+ ...\)
Now look for a pattern in the sums of various number of terms.
i.e. Sum of first 2 terms = 1/2  1/4 = 1/4 Sum of first 3 terms = 1/4 + 1/8 = 3/8 Sum of first 4 terms = 3/8  1/16 = 5/16 Sum of first 5 terms = 5/16 + 1/32 = 11/32 and so on...
You can see that all these sums are between 1/4 and 1/2.
On the GMAT when you have tested such a question for about half the terms and you have established a pattern you can safely assume it will continue. This is because this series is analogous to how the GMAT itself adapts the difficulty of its questions based on your answers. It sort of zigzags in a diminishing pattern. So as you increase k, you will be varying only slightly around a sort of ultimate stagnant value at infinity which you could determine as \(\frac{a}{(1  r)}\) i.e. the sum of an infinite Geometric Progression with first term a and common ratio r.
Here we have first term 1/2 and common ratio 1/2
=> Infinite sum = \(\frac{1}{2*(1  (1/2))}= \frac{1}{2*1.5} = \frac{1}{3}\)
With a large number of terms such as 10 terms, your series sum would actual TEND to stagnate around this infinite sum.
Pick D



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Re: Sequence problem GMATPrep first 10 terms
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18 Aug 2010, 00:17
Initially, I thought about calculating every term. This is almost always the wrong approach. It is more of a number properties problem than sequences.
For the first term, it alternates between positive and negative. For even k, it is positive 1 * \(1/2^k\) and negative 1 for odd k.
The first term is \(1 * 1/2 = 1/2\)
The second term is \(1 * 1/4 = 1/4\)
\(1/2  1/4 = 1/4\)
The third term is \(1 * 1/8 = 1/8\)
\(1/4 + 1/8 = 3/8\)
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 \(1/4\).



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Re: Sequence
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01 Nov 2010, 11:41
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.



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Re: Sequence
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26 Feb 2011, 21:46
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. Using some keen observation, you can quickly arrive at the answer... Terms will be: \(\frac{1}{2}  \frac{1}{4} + \frac{1}{8}  \frac{1}{16} + \frac{1}{32}  ...  \frac{1}{1024}\) For every pair of values: \(\frac{1}{2}  \frac{1}{4} = \frac{1}{4}\) \(\frac{1}{8}  \frac{1}{16} = \frac{1}{16}\) etc... So this series is actually just \(\frac{1}{4} + \frac{1}{16} + ... + \frac{1}{1024}\) So the sum is definitely greater than 1/4. When you add an infinite GP with 1/16 as first term and 1/4 as common ratio, the sum will be \(\frac{\frac{1}{16}}{1\frac{1}{4}} = 1/12\). Here, the sum of terms 1/16 + 1/64 + ... 1/1024 is definitely less than 1/12. Hence the sum is definitely less than 1/2. Answer is (D).
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Re: Sequence
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13 Aug 2011, 12:42
Hello! I was just doing this problem and though I would add some "graphic" approach in case it is useful for someone as it has been for me. You can easily draw a graph of the first points on any series to see if they follow a regular pattern. This one specifically is extremely easy to catch as you draw 3+ points, something like this:
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Re: Sequence
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19 Aug 2011, 16:31
I liked Graphical approach is a good one. BUT  felt there is no need of any calculations T = +(1/2 + 1/2^3 + 1/2^5 + 1/2^7 +1/2^9) (1/2^2 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10)Subtract the top and bottom T = 1/4 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10Now I know T is greater than 1/4 , but if sum of the remaining is smaller than 1/4, then it will definitely be between 1/4 and 1/2 T = 1/4 + 1/4(1/2^2 + 1/2^4 + 1/2^6 + 1/2^8)So sum of the remaining is smaller than 1/4, so ans. is between 1/4 and 1/2 OA D
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Re: Integer K from 1 to 10
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12 Jan 2012, 04:35
enigma123 wrote: For any integer k from 1 to 10, inclusive, the kth of a certain sequence is given by [(1)^(k+1)] × (1 / 2^k). If T is the sum of the first 10 terms of 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 ½ E. less than 1/4
Can someone please let me know what is the concept behind this and how to solve? Every odd value of k will make (k+1) even and [(1)^(k+1)] will be +ve so we will have k = 1, [(1)^(1+1)] × (1 / 2^1) = 1*(1/2) = 1/2 k = 2, [(1)^(2+1)] × (1 / 2^2) = 1*1/4 = 1/4 k = 3, [(1)^(3+1)] × (1 / 2^3) = 1 * 1/8 = 1/8 so we can observe we will get alternate +ve and ve value for (1/2)^k (1/2)  (1/4) + (1/8)  (1/16) + (1/32)  (1/64) + (1/128)  (1/256) + (1/512)  (1/1024) [(1/2)+ (1/8) + (1/32)+ (1/128) + (1/512)]  [(1/4)+(1/16)+(1/64)+(1/256)+(1/1024)] = approx 0.3 so D



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Re: Sequence
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26 Apr 2012, 07:55
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. Other solutions at: sequencecananyonehelpwiththisquestion88628.html#p668661 \(\frac{first \ term}{1constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)?
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Re: Sequence
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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
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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
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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
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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
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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
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Re: Numbers , Squences , Indices
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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
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23 Apr 2013, 00:51
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
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Re: Numbers , Squences , Indices &nbs
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