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Algebra Difficulty 2.8 Junior Find the answer

A sequence has 101 terms, each of which is a positive integer. If a term, nn, is even, the next term is equal to 12n+1\frac{1}{2}n+1. If a term, nn, is odd, the next term is equal to 12(n+1)\frac{1}{2}(n+1). If the first term is 16, what is the 101st term?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The 1st term is 16. Since 16 is even, the 2nd term is 1216+1=9\frac{1}{2} \cdot 16+1=9. Since 9 is odd, the 3rd term is 12(9+1)=5\frac{1}{2}(9+1)=5. Since 5 is odd, the 4th term is 12(5+1)=3\frac{1}{2}(5+1)=3. Since 3 is odd, the 5th term is 12(3+1)=2\frac{1}{2}(3+1)=2. Since 2 is even, the 6th term is 122+1=2\frac{1}{2} \cdot 2+1=2. This previous step shows us that when one term is 2, the next term will also be 2. Thus, the remaining terms in this sequence are all 2. In particular, the 101st term is 2.

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