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

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Joined: 02 Sep 2009
Posts: 46284
In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2 [#permalink]

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02 Feb 2015, 07:19
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Question Stats:

38% (02:14) correct 62% (02:23) wrong based on 235 sessions

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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.

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Posts: 5930
Re: In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2 [#permalink]

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

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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

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Posts: 74
Re: In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2 [#permalink]

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 46284
Re: In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2 [#permalink]

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09 Feb 2015, 04:19
1
5
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.
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Re: In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2 [#permalink]

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

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24 Sep 2017, 05:49
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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
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# In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n>1. Is a3>a2

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