GMAT Prep ?? (Kth term)

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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

Calculate until you see the pattern.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (-1/16, for example).
In other words, the sum will alternate between going up a little and then down a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

Please can anyone explain in detail ... I am not able to grasp the explanation

loki.gmat wrote:here is the shortcut method.
Its a geometric progression for sure.
1st term = 1/2
2nd tern = -1/4
3rd term = 1/8
common ratio = (-1/4)/(1/2) = -1/2

sum of n terms in a geometric progression = a(1-r^n)/(1-r)
where a = 1st term r=common ration n=no of terms

Sn = (1/2)[1-(-1/2)^10]/[1-(-1/2)]
which is approximately equal to 1/3.
hence IMO C.


Thanks!

1/3 ~ 0.33

0.25<0.33<0.5 so, D

GMATGuruNY 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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

Calculate until you see the pattern.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (-1/16, for example).
In other words, the sum will alternate between going up a little and then down a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

Hi Mitch- I like all your explanations. Thanks for all your help. With regard to this question, I reached till the end. i.e. 1/4+1/16+1/64+1/256+1/1024. But how did you conclude that the sum would be somewhere around 1/4 and 1/2. I couldn't understand that part. Could you please explain????

pavan.mpv wrote:

GMATGuruNY 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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

Calculate until you see the pattern.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (-1/16, for example).
In other words, the sum will alternate between going up a little and then down a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

Hi Mitch- I like all your explanations. Thanks for all your help. With regard to this question, I reached till the end. i.e. 1/4+1/16+1/64+1/256+1/1024. But how did you conclude that the sum would be somewhere around 1/4 and 1/2. I couldn't understand that part. Could you please explain????

I'm glad that you've been finding my explanations helpful.
For the problem at hand, the sum is easier to see if you visualize it on a number line:



Follow the arrows.
The first term is 1/2.
When we add in -1/4, the sum decreases to 1/4.
When we add in +1/8, the sum increases to 3/8.
When we add in -1/16, the sum decreases to 5/16.

If we continue drawing arrows, the sum will converge to a value somewhere between 5/16 and 3/8.

The correct answer is D.

If this can be done through GP.

1/2(1-(1/2)^10)/(1-(1/2))=(1023/1024)

hi loki:

ur expln is good but see the below working: i am unable to go further:

1/2 (1023/1024)* 2/3.

how to proceed with further cancellation and end up in 1/3

smackmartine wrote:

loki.gmat wrote:here is the shortcut method.
Its a geometric progression for sure.
1st term = 1/2
2nd tern = -1/4
3rd term = 1/8
common ratio = (-1/4)/(1/2) = -1/2

sum of n terms in a geometric progression = a(1-r^n)/(1-r)
where a = 1st term r=common ration n=no of terms

Sn = (1/2)[1-(-1/2)^10]/[1-(-1/2)]
which is approximately equal to 1/3.
hence IMO C.


Thanks!

1/3 ~ 0.33

0.25<0.33<0.5 so, D

Now that is a rather elaborate question but I am glad to start seeing those on my CAT's! Thanks for the explanation! I am not sure if I can pull that off within the timeframe.

[quote="GMATGuruNY"][quote]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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼[/quote]

Calculate until you see the pattern.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (-1/16, for example).
In other words, the sum will alternate between going up a little and then down a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is [spoiler]D[/spoiler].[/quote]