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If an=f(n) for all integers n>0, is a4>a8?

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If an=f(n) for all integers n>0, is a4>a8? [#permalink]

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New post 23 Jan 2015, 07:40
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A
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C
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E

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  75% (hard)

Question Stats:

48% (01:28) correct 52% (01:22) wrong based on 196 sessions

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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]

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Bunuel wrote:
If \(a_n=f(n)\) for all integers n>0, is \(a_4>a_8\)?

(1) \(a_5=13\)
(2) f(x)=9−f(x−1) for all integers x

Kudos for a correct solution.


1) not sufficient.

2) a(5) = f(5) = 9-f(4) --> not sufficient.

1+2) Since f(x) = 9 - f(x-1) ---> f(x)+f(x-1) = 9. Substituting and working with the equations we get that f(4)=f(8) --> a(4)=a(8)

Answer B.

**edited after spotting a procedural error
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Last edited by gmat6nplus1 on 26 Jan 2015, 03:03, edited 1 time in total.

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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]

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New post 23 Jan 2015, 09:41
Issue (value, compare,...) of numeric arrays may be solved if:
_ Know a value
_ Relationship of that value with others.

Here: a5 =15, a5 = 8 - a4. So, can solve to get a4 and a8 -> Can compare.

Answer C

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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]

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New post 25 Jan 2015, 02:24
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can anyone explain why it is not B.

B states that f(x) = 9 - f(x-1)

=> f(5)+f(4) = 9 --1
=> f(6)+f(5) = 9 --2
=> f(7)+(6) = 9 --3
=> f(8)+f(7) = 9 --4

solving eqautions 1&2
=>f(6) - f(4) = 0 -- 5

solving 5 & 3
=> f(7)+f(4) = 9 --6

solving 6 & 4
=>f(8) - f(4) = 0

=> f(8) = f(4)
=> a8 = a4

there fore a4 cannot be greater than a8

hence B

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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]

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New post 25 Jan 2015, 20:56
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(B) it is. You dont need value of x to calculate f(8). When calculated f(8) = f(4)

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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]

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New post 26 Jan 2015, 04:43
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Bunuel wrote:
If \(a_n=f(n)\) for all integers n>0, is \(a_4>a_8\)?

(1) \(a_5=13\)
(2) f(x)=9−f(x−1) for all integers x

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

There’s very little information in the question stem, so we turn directly to statement (1). Without an equation for f(n), we have no rule to relate the different terms of the sequence. The value of a5 is therefore insufficient standing alone.

Turning to statement (2), the temptation is to assume insufficiency because no term of the sequence is given or can be determined. And of course the two statements will be sufficient taken together, since they will collectively determine every term of the sequence.

Not so fast, though! Things should perhaps feel a bit “too easy” at this point, so let’s explore statement (2) alone a bit further. Combining statement (2) with the question stem gives \(a_n=f(n)=9-f(n-1)=9-a_{n-1}\).

Now, plug in a value for n – say, for instance, n=5. The equation becomes a5=9−a4. Similarly, we obtain a6=9−a5. Finally, substitute for \(a_5:a_6=9-(9-a_4)=9-9+a_4=a_4\). By the same token, a8=a6. It turns out that, regardless of actual values, every other term of the sequence is equal. And since a4=a6=a8, the answer to the question “is a4>a8?” is “no.” So statement (2) is sufficient standing alone.
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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]

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