Difference between revisions of "2007 AMC 12A Problems/Problem 7"

Revision as of 11:05, 22 November 2007

Problem

Let $a, b, c, d$ , and $e$ be five consecutive terms in an arithmetic sequence, and suppose that $a+b+c+d+e=30$ . Which of $a, b, c, d,$ or $e$ can be found?

$\textrm{(A)} \ a\qquad \textrm{(B)}\ b\qquad \textrm{(C)}\ c\qquad \textrm{(D)}\ d\qquad \textrm{(E)}\ e$

Solution

Let $f$ be the common difference between the terms.

$a=c-2f$
$b=c-f$
$c=c$
$d=c+f$
$e=c+2f$

$a+b+c+d+e=5c=30$ , so $c=6$ . But we can't find any more variables, because we don't know what $f$ is. So the answer is $\textrm{C}$ .

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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
All AMC 12 Problems and Solutions

@@ Line 8: / Line 8: @@
 * <math>a=c-2f</math>
-* <math>b\displaystyle =c-f</math>
+* <math>b=c-f</math>
-* <math>c\displaystyle =c</math>
+* <math>c=c</math>
-* <math>d\displaystyle =c+f</math>
+* <math>d=c+f</math>
 * <math>e=c+2f</math>
-<math>a+b+c+d+e=5c=30</math>, so <math>c=6</math>. But we can't find any more variables, because we don't know what <math>f</math> is. So the answer is <math>\textrm{A}</math>.
+<math>a+b+c+d+e=5c=30</math>, so <math>c=6</math>. But we can't find any more variables, because we don't know what <math>f</math> is. So the answer is <math>\textrm{C}</math>.
 ==See also==
 {{AMC12 box|year=2007|num-b=6|num-a=8|ab=A}}

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Difference between revisions of "2007 AMC 12A Problems/Problem 7"

Revision as of 11:05, 22 November 2007

Problem

Solution

See also