It is currently 19 Oct 2017, 04:23

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128823 [0], given: 12183

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

### Show Tags

02 Feb 2015, 07:19
Expert's post
12
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

37% (02:31) correct 63% (02:21) wrong based on 197 sessions

### HideShow timer Statistics

In the sequence a1,a2,…,an,…, $$a_1=x$$ and $$a_n=y-z*a_{n-1}$$ for all n>1. Is $$a_3>a_2$$?

(1) z > y^2 + 2
(2) x > y/(z+1)

Kudos for a correct solution.
[Reveal] Spoiler: OA

_________________

Kudos [?]: 128823 [0], given: 12183

Math Forum Moderator
Joined: 02 Aug 2009
Posts: 4969

Kudos [?]: 5470 [0], given: 112

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

### Show Tags

02 Feb 2015, 10:36
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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5470 [0], given: 112

Senior Manager
Joined: 28 Feb 2014
Posts: 295

Kudos [?]: 141 [0], given: 133

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

### Show Tags

02 Feb 2015, 11:48
Bunuel wrote:
In the sequence a1,a2,…,an,…, $$a_1=x$$ and $$a_n=y-z*a_{n-1}$$ for all n>1. Is $$a_3>a_2$$?

(1) z > y^2 + 2
(2) x > y/(z+1)

i]Kudos for a correct solution.[/i]

$$a_3=y-z*a_{2}$$
$$a_2=y-z*x$$

reworded, $$y-z*a_{2}$$ $$>$$ $$y-z*x$$?
or
$$x$$ $$>$$ $$y/(z+1)$$ ?

Statement 1: Insufficient as it doesn't even contain x.

Statement 2: Exactly what we're looking for. Sufficient

Kudos [?]: 141 [0], given: 133

Manager
Status: Gmat Prep
Joined: 22 Jul 2011
Posts: 74

Kudos [?]: 63 [0], given: 39

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, 06:07
$$a2=y-z*x$$
$$a3=y-z*(y-z*x)=y-z*y+z^2*x$$
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 $$a3>a2$$

hence B.

Kudos [?]: 63 [0], given: 39

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128823 [1], given: 12183

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

### Show Tags

09 Feb 2015, 04:19
1
KUDOS
Expert's post
5
This post was
BOOKMARKED
Bunuel wrote:
In the sequence a1,a2,…,an,…, $$a_1=x$$ and $$a_n=y-z*a_{n-1}$$ for all n>1. Is $$a_3>a_2$$?

(1) z > y^2 + 2
(2) x > y/(z+1)

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

It’s algebra time!

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.
_________________

Kudos [?]: 128823 [1], given: 12183

Intern
Joined: 31 Aug 2013
Posts: 10

Kudos [?]: 6 [0], given: 1

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

### Show Tags

11 Jul 2016, 06:10
Bunuel wrote:
In the sequence a1,a2,…,an,…, $$a_1=x$$ and $$a_n=y-z*a_{n-1}$$ for all n>1. Is $$a_3>a_2$$?

(1) z > y^2 + 2
(2) x > y/(z+1)

Kudos for a correct solution.

---

a little more calculation intensive method :

Here a3=y-z*a2=y-z(y-z*a1)= y*(1-z) + z^2 * x ( here substituting a1 =x)

a2= y-z*x

a3- a2 = [ y*(1-z) + z^2 * x ] - y-z*x = z (x(1+z)-y).

Now a3- a2 > 0 , then z (x(1+z)-y) > 0 or z > 0 and (x(1+z)-y) >0 ( or x > y/(1+z) )

Now 1) says (1) z > y^2 + 2 or z> 0

and 2) says x > y/(1+z) ( which is what we want)

Hence combining 1 and 2, a3-a2 > 0 or a3 > a2

Kudos [?]: 6 [0], given: 1

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16678

Kudos [?]: 273 [0], given: 0

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, 05:49
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.
_________________

Kudos [?]: 273 [0], given: 0

Re: In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2   [#permalink] 24 Sep 2017, 05:49
Display posts from previous: Sort by