Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.9 Junior Find the answer

Let $a_1,a_2,\cdots,a_k$ be a finite arithmetic sequence with $a_4 +a_7+a_{10} = 17$ and $a_4+a_5+\cdots+a_{13} +a_{14} = 77$ .
If $a_k = 13$ , then $k =$

Pick one

Solution

Note that $a_7-3d=a_4$ and $a_7+3d=a_{10}$ where $d$ is the common difference, so $a_4+a_7+a_{10}=3a_7=17$ , or $a_7=\frac{17}{3}$ .
Likewise, we can write every term in the second equation in terms of $a_9$ , giving us $11a_9=77\implies a_9=7$ .
Then the common difference is $\frac{2}{3}$ . Then $a_k-a_9=13-7=6=9\cdot\frac{2}{3}$ .
This means $a_k$ is $9$ terms after $a_9$ , so $k=18\implies\boxed{B}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.