The infinite (serial) sequence a1, a2, , an, is such that a1 = x, a

From the task we have information about a4 element and that our sequence will be repeat every four elements. So we should know information about first four elements
1) we have information about first element - insufficient
2) we have information about sum of second and third elements - insufficient

1+2) from both statements we have information about first and fourth elements and about sum of second and third elements
if we need to calculate 3 elements or 4 elements this will be enough.
But in our case we need to calculate 98 / 4 = 24 group of first 4 elements + sum of first two elements
as we already know we don't have information about exact meaning of second element
so answer is E

VERITAS PREP OFFICIAL SOLUTION:

As people unpack the mystery in this problem, they start to see what’s going on. If an = a(n-4), then each term equals the term that came four prior. So the sequence really goes:

x, y, z, 3, x, y, z, 3, x, y, z, 3…

So although it looks like a pretty massive mystery, really you’re trying to figure out x, y, and z because 3 is just 3. And here’s a common way of thinking:

Statement 1 is not sufficient, but it gets you one of the terms. And Statement 2 is not sufficient but it gets you two more. So when you put them together, you know that the sum of one trip through the 4-term sequence is 5 + 2 + 3 = 10, so you should be able to extrapolate that to the whole thing, right? Just figure out how many trips through will get you to term 98 and you have it; like the Syed jury, you have the motive and the timeline and the cell phone records and Jay’s testimony, so the answer has to be C. Right?

But let’s interview Sarah Koenig here:

Sarah: The pieces all seem to fit but I’m just not so sure. Statement 2 looks really bad for him. If we can connect those dots for y and z, and we already have x, we should have all variables converted to numbers. Literally it all adds up. But I feel like I’m missing something. I can definitely get the sum of the first 4 terms and of the first 8 terms and of the first 12 terms; those are 10, and 20, and 30. But what about the number 98?

And that’s where Sarah Koenig’s trademark thoughtfulness-over-opinionatedry comes in. There is a giant hole in “Answer choice C’s case” against this problem. You can get the sequence in blocks of 4, but 98 is two past the last multiple of 4 (which is 96). The 97th term is easy: that’s x = 5. But the 98th term is tricky: it’s y, and we don’t know y unless we have z with it ( we just have the sum of the two). So we can’t solve for the 98th term. The answer has to be E – we just don’t know.

Now if you’ve heard yesterday’s episode, think about Dana’s “think of all the things that would have to have gone wrong, all the bad luck” rundown. “He lent his car and his phone to the guy who pointed the finger at him. That sucks for him. On the day that his girlfriend went missing. That’s awful luck…” And in real life she may be right – that’s a lot of probability to overcome. But on the GMAT they hand pick the questions. On this problem you can solve for the 97th term (up to 96 there are just blocks of 4 terms, and you know that each block sums to 10, and the 97th term is known as 5) or the 99th term (same thing, but add the sum of the 98th and 99th terms which you know is 2). But the GMAT hand-selected the tricky question just like Koenig hand-selected the Adnan Syed case for its mystery. GMAT Data Sufficiency questions are like Serial…it pays to be skeptical as you examine the evidence. It pays to think like Sarah Koenig. Unlike Jay, the statements will always be true and they’ll always be consistent, but like Serial in general you’ll sometimes find that you just don’t have enough information to definitively answer the question on everyone’s lips. So do your journalistic due diligence and look for alternative explanations (Don did it!). Next thing you know you’ll be “Stepping Out!!!” of the test center with a high GMAT score.

Fell for the trap, chose C as the right answer.

Got the sequence and then calculated 98/4 = 24,5;
24 * (x+y+z+3) + 0,5 * (x+y+z+3) and so on.

however it is not a 100% clear to me, that a98 cannot be derived from the given information. I know that a97+a98+a99+a100 must be equal to 5+2+3 = 10, and two of the parts are 5 and 3. 10 - (5+3) is of course 2, so a98 must be equal to 1?

Hello noTh1ng

You can simplify this task. We have information about sum of repeated \(4\) elements. So we can count any number that multiple of \(4\).
\(4 * 24 = 96\) We can calculate sum of \(96\) numbers.
But we need calculate \(98\) numbers. So we need calculate two first elements.

Can you calculate their sum from given information?

And about your question:
"I know that a97+a98+a99+a100 must be equal to 5+2+3 = 10, and two of the parts are 5 and 3. 10 - (5+3) is of course 2, so a98 must be equal to 1?"
Here is problem: "10 - (5+3) is of course 2" this number \(2\) is made from \(y\) and \(z\) and it can be \(1 + 1\) or \(0 + 2\) or \(2 + 0\) and for our task we need exact value of \(y\) and as you see it can be \(1\), \(0\) or \(2\).

From the given information we see that:-
a5=x
a6=y
a7= z
a8=4

Sequence is repeating after every 4 terms.

That means in 98 terms the sum is 24x+24y+24z +4*24 +x +y= 25x+25y+ 24z+ 4*24
In order to find average, we need to know the value of x, y and z

(1) x = 5
Not sufficient as we don't know the value of y and z

(2) y + z = 2
Not sufficient as we don't know the value of x

Combining both statements and putting in below equation, we still get 'y' left as unknown

25x+ 24(y+z) + y +4*24

E is the answer

So now a1 = x, a2 = y, a3 = z and a4= 3, series will be repeated like this till the 96th term, and we will have a97 = x and a98 = y

From 1, no information about y and z, out

from 2, no information about x, out

Combination would have been good, if we had an even value for x y or z

But that is not the case, said that y and z can take any value, such as 4,-2 or 3,-1

This will give us different values each time

E

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