Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

19 Jan 2013, 11:53

6

This post received KUDOS

22

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

34% (04:56) correct
66% (02:39) wrong based on 832 sessions

HideShow timer Statistics

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

So, we need to find the least value of n for which \(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\) --> \(2^n>7500\) --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

19 Jan 2013, 12:23

Bunuel wrote:

So, we need to find the least value of n for which \(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\) --> \(2^n>7500\) --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

11 Oct 2013, 13:06

Though I got it wrong due to wrong approximation, I found an alternate method.

Take n=1, so G(1) - G(2) = 5(-1/2)^1 - 5(-1/2)^2. On solving we get some number upon power of 2 as 2( since 2^2 is LCM)

Since the value compared is with 1000, we need to take values of 2 above index 10

Take option C: | G(12) - G(13) | = 5(-1/2)^12 - 5(-1/2)^13 , solve it to get 5/2^13. If we take split denominator as 2^3 and 2^10 we get (5/2^3)*(1/2^10).

5/8 = 0.625 and 1/2^10 = 1024 , if you compare it with RHS 1/1000 we are almost there. I had approximated 1024 as 1000, and wrongly took 5/8 as 0.125 which gave me it less than 1000, but in actuals its more than 1/1000 So next number could be the answer, check for G13 and G14, you have the answer. Hope I am able to express it clearly

Honestly speaking I dont think such questions can come up in GMAT

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

02 Jul 2014, 20:01

Bunuel wrote:

maglian wrote:

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

So, we need to find the least value of n for which \(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\) --> \(2^n>7500\) --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

31 Aug 2014, 07:50

Bunuel wrote:

maglian wrote:

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

So, we need to find the least value of n for which \(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\) --> \(2^n>7500\) --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Answer: D.

Hi Bunuel, I got answer C, and also backed it up by calculation with a calculator. it comes down to |-5*(1/2)^n +5*(1/2)*(n+1)| = |-5*(1/2)^(n+1)| < 1/1000 if (n+1) = 13, we get the answer we were looking for, so the answer should be C. Are you sure that the answer needs to be D? Can you find something wrong with my logic/calculations?

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

So, we need to find the least value of n for which \(\frac{15}{2}*(\frac{1}{2})^n<\frac{1}{1000}\) --> \(2^n>7500\) --> the least n for which this inequality hods true is 13 (2^13=~8000>7500).

Thus, the first two terms, g(n) and g(n+1), for which the given inequality holds true are g(13) and g(14).

Answer: D.

Hi Bunuel, I got answer C, and also backed it up by calculation with a calculator. it comes down to |-5*(1/2)^n +5*(1/2)*(n+1)| = |-5*(1/2)^(n+1)| < 1/1000 if (n+1) = 13, we get the answer we were looking for, so the answer should be C. Are you sure that the answer needs to be D? Can you find something wrong with my logic/calculations?

You cannot factor out -1 the way you did. (-1/2)^n does not equal to -1*(1/2)^n if n is even, and we don't know whether it's even or odd.

The whole expression comes down to \(\frac{15}{2}*(\frac{1}{2})^n\).

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

14 Jun 2015, 05:39

maglian wrote:

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

the sequence is 15/4, 15/8, 15/16,......,15/2048, 15/4096, 15/ 8192, 15/16384 and so on so the answer is (D)

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

15 Feb 2016, 00:42

In the last step how did you come to the figure 2^13? how do we arrive at 2 raised to the power of such a big number? esp the choice gives us close options

Re: A sequence of numbers (geometric sequence) is given by the [#permalink]

Show Tags

29 Aug 2016, 02:08

A sequence of numbers (geometric sequence) is given by the expression: g(n)=5 *(-1/2)^n. If the sequence begins with n = 1, what are the first two terms for which |g(n) - g(n+1)| < 1/1000 ?

A. g(10), g(11) B. g(11), g(12) C. g(12), g(13) D. g(13), g(14) E. g(14), g(15)

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...