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Bunuel
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zhanbo
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GMAT 1: 760 Q50 V42
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GMAT 2: 760 Q50 V42
GRE 1: Q169 V168
GRE 2: Q170 V170
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Hi cosmicomet,

This is a genuinely sharp DS instinct. You're right that the algebra in zhanbo's solution quietly divides by `r^11`, and dividing by something that could be zero is exactly the kind of move worth challenging. So credit where it's due.

Here's why it still lands on B, though.

Does `r = 0` even fit? Plug it in: the 12th term would be `7 · 0^11 = 0`, and the 15th term `7 · 0^14 = 0`. Then "15th = 64 × 12th" reads `0 = 64 · 0` - technically true. So mathematically `r = 0` does satisfy the equation.

Why we don't count it. The sequence is built by multiplying by a constant ratio `r` - that's a geometric sequence, and by standard convention its ratio is nonzero. If `r = 0`, the sequence collapses to `7, 0, 0, 0, ...` and there's no real "ratio" left to speak of; the relationships become degenerate `0 = 0` identities that hold for any multiplier. The GMAT treats that as outside the intended setup.

So once `r = 0` is off the table:

- Statement (1): `r^2 = 16` → `r = 4` or `r = -4` → not sufficient.
- Statement (2): `r^3 = 64` → `r = 4` only (one real cube root) → sufficient.

That leaves B. Keep that zero-check habit, though - most DS questions reward it; this one just defines it away.

Answer: B
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