I tried other values, such as

t=3

3/4* 48= 36

t=10

(10/11) * 39= 35 ( the number is decreasing )

So, I’m not getting how @gregmat solve it

I can’t find the video what is the actual answer to the question? Is it QB is greater?

I think the question is testing the subject of SUM in an arithmetic sequence. So I tried to identify the first and the last term within set S, n, and n50.

n1 = 1 - (1/2) = 1/2

n50 = 1 - (1/50) = 49/50

The total # of n = 50

Then, I tried solving it by SUM of arithmetic sequence formula n/2 (n1 + n50) which is:

(50/2) * (37/25) = 37 which is lesser than 42 therefore answer B.

However, I’m not confident with my answer at all, I hope someone could correct me and point out my mistake.

The answer is A. There are a couple of ways you could go around this. The rigorous (but out of scope!) way would be to convert to an integral:

Then make use of the Euler’s theorem - we replace \epsilon (that’s a little over a half) with 2/3 for simplicity:

With the simplification that \ln 50 < 5,

Plugging it back,

and hence the answer is A.

The problem is that this is *not* a solution suitable for the GRE for obvious reasons, and in fact I am not sure of a rigorous alternative either. GregMat’s solution is basically to assume every term to be 6/7 - the reason I am not comfortable with that solution is because he is assuming that

You could show that

and hence try to show that Greg’s relation is true, but it’s a bit more hand-wavy than I like.

we can’t do this because the difference between two terms is not even. This rule applies only for evenly spaced terms.

@gregmat if you could help pls

I just used logic to solve it. It wasn’t a foolproof method, but seemed accurate