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1993 AMC 12 #21 - Arithmetic Sequence

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January 3, 2012
arithmetic sequenceAMC

Problem Statement

Let a1,a2,...,aka_1, a_2, ..., a_k be a finite arithmetic sequence with a4+a7+a10=17 a_4+a_7+a_{10}=17 and a4+a5+a6+a7+a8+a9+a10+a11+a12+a13+a14=77 a_4+a_5+a_6+a_7+a_8+a_9+a_{10}+a_{11}+a_{12}+a_{13}+a_{14}=77 If ak=13a_k=13, then k=k=
<spanclass=′latex−bold′>(A)</span> 16<spanclass=′latex−bold′>(B)</span> 18<spanclass=′latex−bold′>(C)</span> 20<spanclass=′latex−bold′>(D)</span> 22<spanclass=′latex−bold′>(E)</span> 24 <span class='latex-bold'>(A)</span>\ 16 \qquad<span class='latex-bold'>(B)</span>\ 18 \qquad<span class='latex-bold'>(C)</span>\ 20 \qquad<span class='latex-bold'>(D)</span>\ 22 \qquad<span class='latex-bold'>(E)</span>\ 24
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