1993 AHSME Problems/Problem 21

Problem

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 $k =$

$\text{(A) } 16\quad \text{(B) } 18\quad \text{(C) } 20\quad \text{(D) } 22\quad \text{(E) } 24$

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 $k=18\implies\boxed{B}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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