"Just Average"
Q:
a1, a2, a3, ..., an, ...
In the sequence above, each term after the first is equal to the average of the preceding term and the following term.
Quantity A
a10 – a8
Quantity B
a5 – a3
A:
(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.
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If I understand the problem correctly, the second term in the series is the average of the first and third terms. We're not given what the first and third terms are, so I have no idea whether the progression goes in a positive direction (e.g., 1, 2, 3...) or in a negative direction (e.g., -1, -2, -3...). Without knowing this-- and it's crucial to know the value of the third term in relation to the first term-- it seems to me that it's impossible to determine which quantity is greater-- A or B.
ReplyDeleteSo I pick D as my answer.
Is my reasoning faulty?
(Consider, too, that a series like [0, 0, 0, 0...] would satisfy the requirement that every term beyond the first be the average of the terms on either side of it. Positive progression, negative progression, no progression at all... we simply don't have enough information. Or so I think.)
Hmm. My first instinct was C, because it doesn't matter whether the progression is positive or negative--a negative progression will just give a negative answer, but the two will still be equal.
ReplyDeleteI mean, let's just assume that a1 is -1 and a2 is -2, etc. Quantity A will be -2, and so will Quantity B.
I decided to test this out by picking two random numbers to start the sequence. (We don't actually need to know the third term, because if we have the first two everything else follows.) Just to make things interesting, I went with -5 for a1 and 23 for a2 (there are no rules saying that the numbers have to be in order). Starting with these two numbers, we can figure out the rest of the sequence. Through a10, it is as follows:
-5, 23, 18, 13, 8, 3, -2, -7, -12, -17
a10 is -17 and a8 is -7, making Quantity A -10.
a5 is 8 and a3 is 18, making Quantity B -10.
I suspect that this will work for any sequence of numbers possible, so I'm going with C.
There's probably a flaw in my reasoning somewhere, but once again, I'm just brute forcing it. I have no idea what the logic behind this is, or if my reasoning (such as it is) is correct.
I do believe you're right.
ReplyDeleteI think I misunderstood what exactly was meant by Quantities A and B. In my mind, I was thinking that a9, the average of a8 and a10, would have to be greater than a4, the average of a3 and a5, assuming a positive progression. But that's not what Quantities A and B are: they're the differences of two terms separated by a third, and those differences are indeed always the same, no matter how the series progresses, and even if the series is merely a series of zeroes.
ReplyDeleteWell, I won't be winning any GRE study manuals this week, that's for sure. Shame on me.
To put it another way: if a2 is equidistant from a1 and a3 on a number line, and a3 (and all successive terms) are equidistant from their preceding and subsequent terms, then the distance between a1 and a2 does indeed set the tone for the entire series, and that distance between terms is always the same for the entire series. C has to be the answer.
ReplyDeleteHeh. I just noticed that I screwed up the beginning of my series: if the first two terms are -5 and 28, the third term should be 51. I have no idea what I was thinking there, but I managed to get it right from the third term on (that is, if you just drop the first term), which is why I ended up getting the right answer anyway.
ReplyDeleteYour explanation makes sense, though. I don't know if I would have been able to articulate it that way. I was just going on gut and brute force.