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

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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 23 Dec 2012, 23:32
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] New post 23 Dec 2012, 23:39
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fozzzy wrote:
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...


I find the GP formula quite handy.
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Re: Numbers , Squences , Indices [#permalink] New post 22 Apr 2013, 23:47
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 = 1/2-1/4+1/8-1/16....

so it is essentially the sum of differences between two adjcent terms((1/2-1/4)+(1/8-1/16) 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] New post 23 Apr 2013, 01:01
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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


\(S = (\frac{1}{2}+\frac{1}{2^{3}}+..\frac{1}{2^{9}}) - (\frac{1}{2^{2}}+..\frac{1}{2^{10}})\)

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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Last edited by mau5 on 24 Apr 2013, 11:54, edited 2 times in total.
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Re: Numbers , Squences , Indices [#permalink] New post 23 Apr 2013, 02:57
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 1/8-1/16 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] New post 23 Apr 2013, 03:11
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 1/2-1/4+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] New post 23 Apr 2013, 03:13
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] New post 24 Apr 2013, 11:21
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] New post 24 Apr 2013, 11:47
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] New post 11 Oct 2013, 12:22
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] New post 28 Feb 2014, 22:18
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] New post 13 Mar 2014, 07:17
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: Exponents / powers (2) [#permalink] New post 13 Mar 2014, 07:42
Expert's post
chrish06 wrote:
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


Merging similar topics. please refer to the discussion above.
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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 11 Apr 2014, 23:02
Sequence will be 1/2-1/4+1/8-1/16.......

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] New post 05 Nov 2014, 21:58
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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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 06 Nov 2014, 05:46
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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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For every integer k from 1 to 10, inclusive, the kth term of a c [#permalink] New post 20 Apr 2015, 17:42
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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:
[Reveal] Spoiler:
D


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Re: For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 10 May 2015, 18:00
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 [#permalink] New post 11 May 2015, 10:17
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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   [#permalink] 11 May 2015, 10:17

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