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AHSME 1993 Problem 21

    Let a1,a2,⋯ ,aka_1,a_2,\cdots,a_k be a finite arithmetic sequence with a4+a7+a10=17a_4 +a_7+a_{10} = 17 and a4+a5+⋯+a13+a14=77a_4+a_5+\cdots+a_{13} +a_{14} = 77. If ak=13a_k = 13, then k=k = (A) 16(B) 18(C) 20(D) 22(E) 24\text{(A) } 16 \text{(B) } 18 \text{(C) } 20 \text{(D) } 22 \text{(E) } 24

    Dataset text NuminaMath-1.5. Quoted with source.

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