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

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== Solution ==
 
== Solution ==
 
<math>\fbox{B}</math>
 
<math>\fbox{B}</math>
why?
 
  
 
== See also ==
 
== See also ==

Revision as of 16:43, 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

$\fbox{B}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
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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