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The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

This problem is much easier if we can “decode” the sequence Sn = (Sn-1 – 1)2

In other words, Sn is just any term in the sequence (for instance, S3 would be the 3rd term, S10 would be the 10th term, etc.)

Sn-1 just means the term that comes right before Sn.

Therefore, we can rephrase Sn = (Sn-1 – 1)2 as “To get any term, take the term before it, subtract one, then square.”

In this problem, however, we’ve been given S5 and we want S3 – that is, we must go backwards.

To “go backwards” in a sequence, do the opposite of each step, in the opposite order. That is, if, to go from S1 to S2 you would:
1) subtract 1
2) square

…then to go backwards, you would:

1) square root

So, if S5 = 100, square root to get 10 and add 1 to get 11. Notice that we do not have to worry about the possibility of a negative square root, because every term is the square of some number, so no term can be negative.

If S4 = 11, square root to get √11 and add 1 to get √11 + 1. The answer is D.

The problem could also be solved a bit more algebraically as follows:

Sn = (Sn-1 – 1)2

S5 = 100

Therefore:

100 = (S4 – 1)2
10 = S4 – 1 (Again, we drop the possibility of a negative root, because S4 itself is a square.)
S4 = 11

Now repeat the process:

11 = (S3 – 1)2
√11 = S3 – 1
S3 = √11 + 1

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