Sequence with algebra Yikes!

[oquiella]

Please repost with right formatting if possible.

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

(1) z>y2+2
(2) x>yz+1


Please breakdown the algebra.

[manik11]

Please repost with right formatting if possible.

In the sequence a1,a2,...,an,...,a1=x and an=y-Za(n−1) for all n>1. Is a3>a2?

(1) z>y2+2
(2) x>y/(z+1)


Please breakdown the algebra.

I have done some correction in statement 2

So..here goes..
First simplify the question a bit. It asks whether a3>a2
a2 = Y - Za1 --> Y - ZX (a1 = X)
a3 = Y - Za2--> Y -Z(Y-Za1) --> Y-Z(Y-Zx) [Substitute a2 in a3]

Question rephrased:
Is : Y-Z(Y-Zx) > Y - ZX
--> -Z(Y-Zx) > -Zx (subtract Y from both sides)
--> Z(Y-Zx) < Zx (multiply by -1)

Final rephrased question : Is Z(Y-Zx) <Zx

1) z>y^2+2

You can see we can't really deduce anything about our rephrased question from this statement. INSUFFICIENT

2) x>y/(z+1)

Any way you dice it this statement doesn't really help us answer the question either. INSUFFICIENT

Combining 1 and 2

Statement 1: z>y^2+2 From this you know that Z is a number(Y) squared plus 2 so this means that Z must have to be a positive number

With this piece of info lets check out our rephrased question again

rephrased question : Is Z(Y-Zx) <Zx ?

Now since we know the sign of Z (+ ive) we can divide both sides of the inequality by Z
SO..now we've got Is Y-Zx < x ?

rearrange the like terms
Y < X (1+Z)
Divide both sides by (1+Z)....remember we know that Z is positive so Z+1 is positive too. We can divide by (Z+1) without flipping the inequality

Y/(1+Z) < X --->> This is exactly what statement 2 tells us. Awesome!

Final answer : C

[Matt@VeritasPrep]

Please repost with right formatting if possible.

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

(1) z>y2+2
(2) x>yz+1


Please breakdown the algebra.

I'm not sure what your formatting should be: is it

a� = x
aₑ = y - z*aₑ₋�

?

Not sure how to read S2 either.