Difference between revisions of "1993 AHSME Problems/Problem 21"

Latest revision as of 16:49, 26 August 2017

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 $k =$

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

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

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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 This means <math>a_k</math> is <math>9</math> terms after <math>a_9</math>, so <math>k=18\implies\boxed{B}</math>
 == See also ==
 {{AHSME box|year=1993|num-b=20|num-a=22}}