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 S_n = (S_n-1 – 1)²

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

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

Therefore, we can rephrase S_n = (S_n-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
2) add 1

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:

S_n = (S_n-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

The correct answer is D.