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For every integer k from 1 to 10, inclusive the "kth term of a certain

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 24 Apr 2013, 12:37
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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
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 25 Apr 2013, 00:55
1
TheNona wrote:
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 :)


Check out the GP perspective on this question too. It really cuts down your work:
http://www.veritasprep.com/blog/2012/04 ... rspective/
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Re: For every integer k from 1 to 10, inclusive, the kth term of a certain  [#permalink]

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New post 24 Feb 2014, 18:23
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I did it the following way.

K=(1/2)-(1/4)+(1/8)-(1/16)+..... ----- 1

Multiply K by 2

2K=1-(1/2)+(1/4)-.....-(1/512) ------- 2

Adding 1 and 2

3K = 1 -(1/1024)
K= (1/3) -{1/(3*1024)}

Now 1/(3*1024) will be very small
So K= 1/3 = .3333

Ans Option D
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Re: For every integer k from 1 to 10, inclusive, the kth term of a certain  [#permalink]

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New post 12 Jul 2014, 07:54
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This is GP.

The terms will be 1/2-1/4+1/8-....

Common ratio is (-1/4)/(1/2) = -1/2

So the sum of terms = 1/2 [1- (-1/2)^10]/(1-(-1/2)) = 1/2 *[1-1/1024]/3/2 = 1023/(1024*3) close to 1/3 so Option D
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Re: For every integer k from 1 to 10, inclusive, the kth term of a certain  [#permalink]

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New post 04 Aug 2014, 06:45
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1
http://tinypic.com/r/2lvk4z5/8

Please check a simpler solution to the above problem in the above image link.

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 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 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 \(sum=\frac{b}{1-r}\), 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: sequence-can-anyone-help-with-this-question-88628.html#p668661


Alternative, if you use the geometric series formula.

S = \frac{a(1-r^n)}{1-r}

where a = first term, r = multiple factor, n = # of terms.

Hi Bunuel, how are these two formula different? Thank you.
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New post 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 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 \(sum=\frac{b}{1-r}\), 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: sequence-can-anyone-help-with-this-question-88628.html#p668661


Alternative, if you use the geometric series formula.

S = \frac{a(1-r^n)}{1-r}

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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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 20 Apr 2015, 18:42
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Hi All,

As complex as this question looks, it's got a great pattern-matching 'shortcut' built into it. When combined with the answer choices, you can avoid some of the calculations....

By plugging in the first few numbers (1, 2, 3, 4), you can see that a pattern emerges among the terms....

1st term = 1/2
2nd term = -1/4
3rd term = 1/8
4th term = -1/16

The terms follow a positive-negative-positive-negative pattern all the way to the 10th term and each term is the product of the prior term and 1/2. By "pairing up' the terms, another pattern emerges....

1/2 - 1/4 = 1/4

1/8 - 1/16 = 1/16

1/32 - 1/64 = 1/64

Etc.

The "pairs" get progressively smaller (notice how each is the product of the prior term and 1/4). This means that we're "starting with" 1/4 and adding progressively TINIER fractions to it. Since we're just adding 4 progressively smaller fractions to 1/4, this means that we're going to end up with a total that's just a LITTLE MORE than 1/4. Looking at the answer choices, there's only one answer that fits:

Final Answer:

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 10 May 2015, 19:00
I've never seen this term "geometric progression" in my studies thus far - is there a good overview of them somewhere and potential questions that might be asked in reference to them? Thanks!
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 10 May 2015, 21:07
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1
healthjunkie wrote:
I've never seen this term "geometric progression" in my studies thus far - is there a good overview of them somewhere and potential questions that might be asked in reference to them? Thanks!



Here is a post that explains Geometric progressions (GP):
http://www.veritasprep.com/blog/2012/04 ... gressions/

The GP perspective on this question is discussed here:
http://www.veritasprep.com/blog/2012/04 ... rspective/
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 11 May 2015, 11:17
Hi healthjunkie,

Geometric progressions are rather rare on the GMAT (while you will see at least 1 sequence question on Test Day, it is not likely to be a Geometric sequence), so you shouldn't be putting too much effort into this concept just yet.

How are you performing on the Quant section overall? How about in the 'big' categories (Algebra, Arithmetic, Number Properties, DS, etc.)? That's where you're going to find the bulk of the points.

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Re: For every integer k from 1 to 10, inclusive, the kth term of a certain  [#permalink]

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New post 03 Jun 2015, 13:56
I used the same calculation as above, which will probably take little more than 2 minutes. Is there a simple version to solve this problem?
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Re: For every integer k from 1 to 10, inclusive, the kth term of a certain  [#permalink]

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New post 03 Jun 2015, 15:14
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Hi vijaydoli,

This question comes up every so often in this Forum. There are a couple of different ways of thinking about this problem, but they all require a certain degree of "math."

Without too much effort, you can deduce what the sequence is:

+1/2, -1/4, +1/8, -1/16, etc.

The "key" to solving this question quickly is to think about the terms in "sets of 2"…

1/2 - 1/4 = 1/4

Since the first term in each "set of 2" is greater than the second (negative) term, we now know that each set of 2 will be positive.

1/8 - 1/16 = 1/16

Now we know that each additional set of 2 will be significantly smaller than the prior set of 2.

1/4....1/16....1/64....etc.

Without doing all of the calculations, we know….
We have 1/4 and we'll be adding tinier and tinier fractions to it. Since there are only 10 terms in the sequence, there are only 5 sets of 2, so we won't be adding much to 1/4. Based on the answer choices, only one answer makes any sense…

Final Answer:

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 12 Oct 2015, 21:33
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VeritasPrepKarishma 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.


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).


Quote:
Hi Karishma

The first term in this example is 1/2. Can you kindly explain how to calculate the sum of all terms of a GP with constant ratio >1 ?

Thanks


Sum of n terms of a GP = a(1 - r^n)/(1 - r)

The formula is the same whether |r| is more than 1 or less than 1.

You can find the sum of an infinite GP by the formula a/(1 - r) only when |r| < 1.
You cannot find the sum of an infinite GP when |r| > 1 because the sum will be infinite.
e.g. 3 + 9 + 27 + 81 ...... infinite terms - The sum will be infinite since you keep adding larger and larger terms.
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New post 15 Jan 2017, 11:48
Seriously - would anyone be able to resolve in 2 minutes?
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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).

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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 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^n-1)/r-1 ?

Please explain..

Bunuel wrote:
Stiv wrote:
\(\frac{first \ term}{1-constant}\) 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 \(sum=\frac{b}{1-r}\), where \(b\) is the first term.
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New post 23 Feb 2017, 00:15
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^n-1)/r-1 ?

Please explain..

Bunuel wrote:
Stiv wrote:
\(\frac{first \ term}{1-constant}\) 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 \(sum=\frac{b}{1-r}\), where \(b\) is the first term.


Infinite progressions are those with infinite number of terms. Whereas a finite sequence has defined first and last terms.
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Re: For every integer k from 1 to 10, inclusive the "kth term of a certain  [#permalink]

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New post 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^n-1)/r-1 ?

Please explain..

Bunuel wrote:
Stiv wrote:
\(\frac{first \ term}{1-constant}\) 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 \(sum=\frac{b}{1-r}\), 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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New post 14 Jun 2017, 16:18
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   [#permalink] 14 Jun 2017, 16:18

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