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:

hi.. ans B.. firstly what does the question ask us.. a1=x.. a2=y-x*z... a3= y-z*(y-x*z).. we have to find whether a3>a2..ie y-z*(y-x*z)>y-x*z, which on simplification becomes is x>y/(1+z).. 1) statement gives relation between two variables not sufficient.. 2) this statement tells us what we have to find x>y/(1+z).. so sufficient
_________________

Re: In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2 [#permalink]

Show Tags

03 Feb 2015, 05:07

\(a2=y-z*x\) \(a3=y-z*(y-z*x)=y-z*y+z^2*x\) in order to have \(a3> a2\)

\(y-z*y+z^2*x>y-z*x\) which after simplifying becomes \(y-z*x<x\) \(y<x+z*x\) \(y<x*(1+z)\) (1) insufficient as it does not tells us about x (2) sufficient as its the expression we need to prove \(a3>a2\)

Using the sequence rule, \(a_2=y-za_1=y-zx\). And \(a_3=y-za_2=y-z(y-zx)\).

With these expressions, the question is a3>a2” becomes “is y–z(y–zx)>y–zx?”

Subtract y from both sides:

“is −z(y–zx)>−zx?”

Divide by -1 (don’t forget to flip the inequality!) to get a cleaner question:

“is z(y–zx)<zx?”

However, we can go no further at this point, since the next simplification would be to divide by z. This is unacceptable, because we don’t know whether z is negative and therefore don’t know whether to flip the inequality again.

Let’s turn now to the additional premises.

Statement (1) seems a bit random, but it does do one important thing for us: It establishes that z is positive. (Recall that even powers are always non-negative, which means that z is greater than a nonnegative number plus a positive number.) This allows us to divide both sides of the question “is z(y–zx)<zx?” by z, resulting in:

“is y–zx<x?”

However, we still have three unknowns and only two unconnected inequalities, so statement (1) is insufficient.

Turning to statement (2), we face the same problem. Knowing that x>y/(z+1) just doesn’t answer the question “is z(y–zx)<zx?” There are too many variables and too few unconnected inequalities. Statement (2) is insufficient.

Finally we combine both statements (1) and (2). Recall that statement (1) simplified the question to “is y–zx<x?” Since statement (2) is solved for x, solve this question for x as well:

“is y<zx+x?” “is y<(z+1)x?” We can divide by z+1 without flipping the inequality since, remember, z is positive: “is x>y/(z+1)?” Now compare statement (2). We know that x>y/(z+1), so… hey, wait a minute… it’s exactly the same! The two statements taken together are sufficient.
_________________

Re: In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2 [#permalink]

Show Tags

24 Sep 2017, 04:49

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________