In the sequence a1,a2,…,an,…,a1=x and an=y–zan−1 for all n1. Is a3a2

Question

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 .

Bunuel · Accepted Answer

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)y/(z+1) just doesn’t answer the question “is z(y–zx)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.

chetan2u · Answer

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

peachfuzz · Answer

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

Answer: B

ynaikavde · Answer

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*xa2 hence B.

abhidpyrate · Answer

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

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