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.

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:

The sequence s1, s2, s3,.....sn,...is such that Sn= [#permalink]
26 Apr 2012, 06:14

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct
0% (00:00) wrong based on 7 sessions

I don't know whether these problems have already been posted on the site, since I couldn't find the answers I will post them.

1) The sequence s_1, s_2, s_3, ..., s_n, ... is such that s_n=1/n - 1/n+1 for all integers n\geq 1. If k is a positive Integer, is the sum of the first k terms of the sequence greater than [fraction]{9}{10}[/fraction]? (1) k > 10 (2) k < 19

2) In the sequence x_0, x_1, x_2, ..., x_n, each term from x_1 to x_kis 3 greater than the previous term, and each term from x_k+1 to x_nis less than the previous term, where n and k are positive integers and k< n. If x_0 = x_n = 0 and if x_k = 15, what is the value of n? A) 5 B) 6 C) 9 D) 10 E) 15

Please elaborate these problems as simple as possible! Thank you! _________________

Kudos if you like the post!

Failing to plan is planning to fail.

Last edited by Bunuel on 26 Apr 2012, 06:54, edited 1 time in total.

Re: Two sequence problems [#permalink]
26 Apr 2012, 06:54

Expert's post

Stiv wrote:

I don't know whether these problems have already been posted on the site, since I couldn't find the answers I will post them.

1) The sequence s_1, s_2, s_3, ..., s_n, .., is such that [m]s_n=1/n - 1/n+1 for all integers n\geq 1. If k is a positive Integer, is the sum of the first k terms of the sequence greater than [fraction]{9}{10}[/fraction]? (1) k > 10 (2) k < 19

2) In the sequence x_0, x_1, x_2, ..., x_n, each term from x_1 to x_kis 3 greater than the previous term, and each term from x_k+1 to x_nis less than the previous term, where n and k are positive integers and k< n. If x_0 = x_n = 0 and if x_k = 15, what is the value of n? A) 5 B) 6 C) 9 D) 10 E) 15

Please elaborate these problems as simple as possible! Thank you!

Re: The sequence s1, s2, s3,.....sn,...is such that Sn= [#permalink]
26 Apr 2012, 06:56

Expert's post

The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - (1/(n+1)) for all integers n>=1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10?

Given: s_n=\frac{1}{n}-\frac{1}{n+1} for n\geq{1}. So: s_1=1-\frac{1}{2}; s_2=\frac{1}{2}-\frac{1}{3}; s_3=\frac{1}{3}-\frac{1}{4}; ...

If you sum the above 3 terms you'll get: s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4} (everything but the first and the last numbers will cancel out). So the sum of first k terms is fgiven by the formula sum_k=1-\frac{1}{k+1}.

Question: is sum_k=1-\frac{1}{k+1}>\frac{9}{10}? --> is \frac{k}{k+1}>\frac{9}{10}? --> is k>9?

(1) k > 10. Sufficient. (2) k < 19. Not sufficient.

Re: The sequence s1, s2, s3,.....sn,...is such that Sn= [#permalink]
26 Apr 2012, 06:57

Expert's post

In the sequence x_0, \ x_1, \ x_2, \ ... \ x_n, each term from x_1 to x_k is 3 greater than the previous term, and each term from x_{k+1} to x_n is 3 less than the previous term, where n and k are positive integers and k<n. If x_0=x_n=0 and if x_k=15, what is the value of n?

A.5 B. 6 C. 9 D. 10 E. 15

Probably the easiest way will be to write down all the terms in the sequence from x_0=0 to x_n=0. Note that each term from from x_0=0 to x_k=15 is 3 greater than the previous and each term from x_{k+1} to x_n is 3 less than the previous term:

So we'll have: x_0=0, 3, 6, 9, 12, x_k=15, 12, 9, 6, 3, x_n=0. So we have 11 terms from x_0 to x_n thus n=10.