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AHSME 1979 Problem 14

    In a certain sequence of numbers, the first number is 11, and, for all n≥2n\ge 2, the product of the first nn numbers in the sequence is n2n^2. The sum of the third and the fifth numbers in the sequence is (A) 259(B) 3115(C) 6116(D) 576225(E) 34\textbf{(A) }\frac{25}{9} \textbf{(B) }\frac{31}{15} \textbf{(C) }\frac{61}{16} \textbf{(D) }\frac{576}{225} \textbf{(E) }34

    Dataset text NuminaMath-1.5. Quoted with source.

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    \frac{61}{16}

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