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# For every integer k from 1 to 10, inclusive, the kth term of

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07 Jan 2010, 05:22
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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
[Reveal] Spoiler: OA

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07 Jan 2010, 06:30
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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 $$|r|<1$$, 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.

Other solutions at: sequence-can-anyone-help-with-this-question-88628.html#p668661
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13 Aug 2011, 11:42
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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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07 Jan 2010, 08:13
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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.

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=(1-1/1024) / 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 problem GMATPrep first 10 terms [#permalink]

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17 Aug 2010, 23:17
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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: DS : Number Theory [#permalink]

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10 Jul 2010, 00:10
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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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01 Nov 2010, 10:41
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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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19 Aug 2011, 15:31
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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 + 1/2^10
Now 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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26 Feb 2011, 20:46
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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: For every integer k from 1 to 10, inclusive, the kth term of [#permalink]

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10 May 2015, 20:07
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7185 Location: Pune, India Followers: 2167 Kudos [?]: 14017 [2] , given: 222 Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] ### Show Tags 12 Oct 2015, 20:33 2 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Integer K from 1 to 10 [#permalink]

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12 Jan 2012, 03:35
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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: Numbers , Squences , Indices [#permalink]

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22 Apr 2013, 23:51
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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 [#permalink]

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24 Apr 2013, 11:37
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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 .

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: Numbers , Squences , Indices [#permalink]

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24 Apr 2013, 23:55
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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 .

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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27 Apr 2010, 18:08
Nice question and nice explanations! I was more leaning toward geometric series solution.
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26 Apr 2012, 06:55
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 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.

Other solutions at: sequence-can-anyone-help-with-this-question-88628.html#p668661

$$\frac{first \ term}{1-constant}$$ Is this formula reversed when we have an increase by 0<constant<1? Does it look like this $$\frac{first \ term}{1+constant}$$?
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26 Apr 2012, 07:07
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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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Re: Sequence   [#permalink] 26 Apr 2012, 07:07

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