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

Revision as of 00:47, 26 September 2014

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 $k =$

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

$\fbox{B}$

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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-Let a_1,a_2,...,a_k be a finite arithmetic sequence with a_4 +a_7+a_10 = 17 and a_4+a_5+...+a_13 +a_14 = 77. if a_k = 13, then k =
+== Problem ==
-(A) 16     (B) 18     (C) 20     (D) 22     (E) 24
+Let <math>a_1,a_2,\cdots,a_k</math> be a finite arithmetic sequence with <math>a_4 +a_7+a_{10} = 17</math> and <math>a_4+a_5+\cdots+a_{13} +a_{14} = 77</math>.
+If <math>a_k = 13</math>, then <math>k =</math>
+<math>\text{(A) } 16\quad
+\text{(B) } 18\quad
+\text{(C) } 20\quad
+\text{(D) } 22\quad
+\text{(E) } 24</math>
+== Solution ==
+<math>\fbox{B}</math>
+== See also ==
+{{AHSME box|year=1993|num-b=20|num-a=22}}
+[[Category: Intermediate Algebra Problems]]
 {{MAA Notice}}