Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 12: Line 12:
  
 
== Solution ==
 
== Solution ==
−
Let d be the common difference. Note that <math>a_7-3d=a_4</math> and <math>a_7+3d=a_{10}</math> where <math>d</math> is the common difference, so <math>a_4+a_7+a_{10}=3a_7=17</math>, or <math>a_7=\frac{17}{3}</math>.Likewise, we can write every term in the second equation in terms of <math>a_9</math>, giving us <math>11a_9=77\implies a_9=7</math>.Then the common difference is <math>\frac{2}{3}</math>. Then <math>a_k-a_9=13-7=6=9\cdot\frac{2}{3}</math>.This means <math>a_k</math> is <math>9</math> terms after <math>a_9</math>, so <math>k=18\implies\boxed{B}</math>
+
Let <math>d</math> represent the common difference. Note that <math>a_7-3d=a_4</math> and <math>a_7+3d=a_{10}</math> where <math>d</math> is the common difference, so <math>a_4+a_7+a_{10}=3a_7=17</math>, or <math>a_7=\frac{17}{3}</math>.Likewise, we can write every term in the second equation in terms of <math>a_9</math>, giving us <math>11a_9=77\implies a_9=7</math>.Then the common difference is <math>\frac{2}{3}</math>. Then <math>a_k-a_9=13-7=6=9\cdot\frac{2}{3}</math>.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 ==
 
== See also ==

Revision as of 16:46, 26 August 2017

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

Let $d$ represent the common difference. 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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_21&oldid=87206"