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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]
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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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(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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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]
Maybe I'm missing something completely obvious here, but how can we use (2) when there is no reference to n? There is no relationship given between x and n, so how can we use a function that is specified "for all integers x" to satisfy an inequality that is defined by n?
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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]
chris74j

per the question stem n>0 and per statement 2, x applies to ALL integers, which means statement 2 applies to positive and negative integers.

Thus, you can use statement 2 equation for all x>0 and x<0, but x>0 integers is basically n as per the question (n>0).

Hope this makes sense
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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]
Bunuel wrote:
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.




Bunuel : In DS the two statements shouldn't contradict each other right?

(1) Says a5 is 13. If we substitute this in the equation,
an+an-1 = 9 ( a6 + a5 = 9 )
a6 = -4

But according to your explanation each term is equal. Arent the two statements contradicting each other?
Sorry if I m missing out on something here!
Please help.
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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]
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KunalR24 wrote:
Bunuel wrote:
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.




Bunuel : In DS the two statements shouldn't contradict each other right?

(1) Says a5 is 13. If we substitute this in the equation,
an+an-1 = 9 ( a6 + a5 = 9 )
a6 = -4

But according to your explanation each term is equal. Arent the two statements contradicting each other?
Sorry if I m missing out on something here!
Please help.


The solution says that "every other term of the sequence is equal". So, \(a_2=a_4=a_6=a_8=...=-4\) and \(a_1=a_3=a_5=a_7=...=13\).
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If an=f(n) for all integers n>0, is a4>a8? [#permalink]
Another approach that I used :


(1) clearly useless for now.

(2) f(n)=9-f(n-1) ***************** 1

f(n-1)=9-f(n-2) *************** 2

replace f(n-1) of 2 in one
f(n)=9-(9-f(n-2))=f(n-2)

so f(8)=f(6)=f(4) then a4 = a8

B sufficient
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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]
B because you can work your way down from f8 to f4 and find out that they're actually equal. So statement 2 is sufficient.
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Re: If an=f(n) for all integers n>0, is a4>a8? [#permalink]
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