Is the sum of the 12th and 13th terms in a sequence of integers odd? (1) Every term in the sequence is 1 more than the preceding term. (2) The third term of the sequence is 4.
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Is the answer A? Because statement 1 tells us it's consecutive integers, so parity alternates, and 12th and 13th would be opposite parity... so sum is odd. Statement 2 doesn't tell us the pattern.
Yeah I got A too. The alternating parity thing clicked once I wrote out a few terms.
Wait, does statement 2 alone really not work? If the third term is 4, can't we figure out the 12th and 13th somehow?
Nope, we don't know if the sequence is arithmetic, geometric, or random. T3=4 tells us nothing about T12 and T13 individually.
Right, that was my trap too. I almost picked C but then realized we still don't know the pattern.
Took me a second but once I realized consecutive integers alternate odd/even, it's straightforward. Solid 500-level question.
Quick tip: for these parity questions, just test with small numbers. If T12 is even, T13 is odd, sum is odd. If T12 is odd, T13 is even, sum still odd. Works every time.
What it tests
Your ability to spot and extend arithmetic and geometric sequences and work with recursive definitions.
Common trap
Off-by-one errors in indexing (a₀ vs a₁) or confusing an arithmetic difference with a geometric ratio.