Sequence S is defined as Sn = X + (1/X), where X = Sn– 1 + 1

Given:
\(S_n = X + \frac{1}{X}\)
\(X = S_{n – 1} + 1\)

This means
\(X = S_{1} + 1 = 201 + 1=202\)
\(S_2 = 202 + \frac{1}{202}\)

So S2 is 1 more than S1 and then a tiny bit extra. S2 is approximately 1 more than S1. Then S3 will be approximately 1 more than S2 and so on. The only substantial increase from one term to the other is 1. The other increase is very very small (of the order of 1/200 which is .005. On the other hand, the options give a range of 1000. Hence, this tiny amount can be easily ignored.

First 50 terms will be
201, 202, 203 .. 250

Sum = Avg * 50 = (201 + 250)/2 * 50 = 451 * 25 = 11275

The actual sum will be a tiny bit more than this.

Answer (C)

Check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... gressions/

Breaking Down the Info:

X can be substituted, so write \(S_n = S_{n - 1} + 1 + \frac{1}{S_{n - 1} + 1}\)

We start with \(S_1 = 201\), then the fraction portion of each term, hence \(\frac{1}{S_{n - 1} + 1}\) , is negligible.

Then we can treat this sequence as \(S_n = S_{n-1} + 1\), which is an arithmetic sequence. The median of the 50 terms would be near \(225.5\). Then the sum is for our estimate is \(225.5*50 = 11250\). We expect the true value to be only a little bigger than that.

Answer: C

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