BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT PREP QUESTION w/OA

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 347
Joined: Mon Aug 04, 2008 1:42 pm
Thanked: 1 times

GMAT PREP QUESTION w/OA

by Stockmoose16 » Sun Oct 26, 2008 3:42 pm
Can someone please explain the following GMAT PREP question? I don't understand the language in the question? Please don't just answer the question -- explain what the question is asking for.


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 1/2
E. less than 1/4

OA is D
Top
Source: — Problem Solving |
Legendary Member
Posts: 541
Joined: Thu May 31, 2007 6:44 pm
Location: UK
Thanked: 21 times
Followed by:3 members
GMAT Score:680

by rohangupta83 » Sun Oct 26, 2008 3:57 pm
equation is [(-1)^(k+1)]*(1 / 2^k)

for k = 1

you will get: [(-1)^2]*(1/2) = 1*1/2=1/2

k=2 you will get (-1)*1/4 = -1/4

k=3 you will get (1)* 1/8 = 1/8

k=4 you will get -1/16

k=5 you will get 1/32

and so on......

Note your series would look something like this

1/2 - 1/4 + 1/8 - 1/16 + 1/32..........

basically, you are subtracting 1/4 from 1/2 you get 1/4

then you add 1/8 to 1/4 = 3/8...

all in all, your end result for the series should be between 1/2 and 1/4
Top
Master | Next Rank: 500 Posts
Posts: 347
Joined: Mon Aug 04, 2008 1:42 pm
Thanked: 1 times

by Stockmoose16 » Sun Oct 26, 2008 4:00 pm
rohangupta83 wrote:equation is [(-1)^(k+1)]*(1 / 2^k)

for k = 1

you will get: [(-1)^2]*(1/2) = 1*1/2=1/2

k=2 you will get (-1)*1/4 = -1/4

k=3 you will get (1)* 1/8 = 1/8

k=4 you will get -1/16

k=5 you will get 1/32

and so on......

Note your series would look something like this

1/2 - 1/4 + 1/8 - 1/16 + 1/32..........

basically, you are subtracting 1/4 from 1/2 you get 1/4

then you add 1/8 to 1/4 = 3/8...

all in all, your end result for the series should be between 1/2 and 1/4
I don't even understand what the questions is asking for. The language is very odd. And how would you do your method in under 2 minutes?
Top
Legendary Member
Posts: 541
Joined: Thu May 31, 2007 6:44 pm
Location: UK
Thanked: 21 times
Followed by:3 members
GMAT Score:680

by rohangupta83 » Sun Oct 26, 2008 4:16 pm
well, the question says that:

For any integer k from 1 to 10, inclusive - which means k could be 1-10 (inclusive)

kth of a certain sequence is given by [(-1)^(k+1)]*(1 / 2^k) - which means that kth term is given by the above sequence

For example you want to find out the first term of the sequence (remember, there are only 10 terms in the sequence with k from 1-10) i.e. k=1

plug k = 1 into the above formual and you get the first term and so on

Finally - the question is asking you to find the sum of first 10 terms of this sequence i.e. find the sum of the sequence given by the formula [(-1)^(k+1)]*(1 / 2^k) where 1<= k <=10

Hopefully this should clear what the question is asking you to do.
Last edited by rohangupta83 on Sun Oct 26, 2008 4:23 pm, edited 1 time in total.
Top
Legendary Member
Posts: 541
Joined: Thu May 31, 2007 6:44 pm
Location: UK
Thanked: 21 times
Followed by:3 members
GMAT Score:680

by rohangupta83 » Sun Oct 26, 2008 4:22 pm
and in the solution, all you need are the first 4 terms at most to give you an idea what approximately is your answer.

For example

lets assume that the first term is 100

you take half out i.e subtract 50 so now you are left with 50

now you are adding 25 to this 50 - you have 75

now subtracting 12.5 from 75 - you have 62.5

and if you continue the same series which is adding half of what you just subtracted to the result and then subtracting half of what you just added from the result, you will notice that you will oscillate between the range of 50-100, never crossing either extremes.

It is the same principle implied in the above problem
Top
User avatar
Legendary Member
Posts: 2134
Joined: Mon Oct 20, 2008 11:26 pm
Thanked: 237 times
Followed by:25 members
GMAT Score:730

by logitech » Sun Oct 26, 2008 11:33 pm
This is an approximation question, hope this helps.
Attachments
prob.jpg
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
Top
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

by sanju09 » Wed Apr 08, 2009 4:28 am
T = 1/2 - 1/2^2 + 1/2^3 - ... -1/2^10

= 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5

Notice that 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 < 1/4, we can say that 1/4 < T < 1/2.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Sat May 30, 2009 4:12 pm
Location: Philadelphia

Approximation question

by bbukoy2 » Sat Jul 25, 2009 12:49 pm
Can someone please clarify how to solve this problem with 2 mins:
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:
logitech wrote:This is an approximation question, hope this helps.
bbuks
Top
Master | Next Rank: 500 Posts
Posts: 113
Joined: Thu Jul 16, 2009 11:23 am
Thanked: 15 times
GMAT Score:730

by prindaroy » Sat Jul 25, 2009 1:54 pm
You can use the formula for a geometric progression if you want.
Top
Master | Next Rank: 500 Posts
Posts: 148
Joined: Wed Jun 03, 2009 8:04 pm
Thanked: 18 times
Followed by:1 members

by ghacker » Mon Jul 27, 2009 2:09 pm
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:

It has two parts (i) (-1)^(k+1) we see that the answer will be +1 when k is odd and -1 when k is even

Why ? (-1) ^even = +1 and (-1)^odd = -1 , but when k = even k+1 will be odd and when k = odd k+1 will be even

(ii) so there will +ve values as well as -ve value

From (i) we see that for even k the given function is -ve and for odd k the given function is +ve

If we add them we get a nice expression ( ref figure)
Attachments
A.JPG
Top
Master | Next Rank: 500 Posts
Posts: 221
Joined: Wed Jan 21, 2009 10:33 am
Thanked: 12 times
Followed by:1 members

by krisraam » Mon Jul 27, 2009 6:13 pm
(1/2 + 1/2^3 +1/2^5....) - 1/2 (1/2 + 1/2^3 + ...)

= 1/2(1/2 + 1/2^3 + ...)
= 1/4 + 1/4( 1/4 + 1/2^4...)
= 1/4 + 1/4( value will bw < 1)

Sum ranges between 1/4 and 1/2

Thanks
Raama
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Tue Jan 06, 2009 1:07 pm

maybe I am wrong, but...

by dwilliams05 » Wed Sep 09, 2009 4:36 pm
when you plug in k=1, you get 1/2. as k increases shouldn't the value of the formula decrease? Hence the answer will never be more than 1/2 so you can eliminate any answer choice that doesn't fit that parameter. Or am I wrong in using that technique?
Top
Post new topic Post Reply
• Page 1 of 1