What we see: a sequence Data Sufficiency question. From the third term on, each term is the sum of the two terms before it. We need the first term.
Step 1. Write the rule both ways.
What we see: \(t_n = t_{n-1} + t_{n-2}\) for \(n \ge 3\).
What it means: the rule also runs backward: \(t_{n-2} = t_n - t_{n-1}\). Any two terms side by side let us walk back to the start.
What we do and why: a statement is enough if it gives two terms that sit next to each other, with their places known.
Step 2. Statement (1): one term is 29.
What we see: one number, and we do not know where it sits.
What it means: many sequences contain 29.
What we do and why: test two. The sequence 29, 1, 30, ... has first term 29. The sequence 1, 28, 29, ... has first term 1. Two different first terms, so statement (1) is not enough.
Step 3. Statement (2): \(t_4 = 11\) and \(t_5 = 18\).
What we see: two terms next to each other, with their places known.
What it means: we can walk backward one step at a time.
What we do and why:
\(t_3 = t_5 - t_4 = 18 - 11 = 7\)
\(t_2 = t_4 - t_3 = 11 - 7 = 4\)
\(t_1 = t_3 - t_2 = 7 - 4 = 3\)
Check forward: 3, 4, 7, 11, 18. It works, and the first term is fixed. Statement (2) is enough.
The trap: picking "both together" because statement (1) seems to add a clue. It adds nothing: the sequence from statement (2) does reach 29 as its sixth term (3, 4, 7, 11, 18, 29), so (1) only agrees with what (2) already fixed.
Takeaway: In a "sum of the two before" sequence, two terms side by side with known places fix every term, forward and backward.
Answer: B
