Difference between revisions of "2007 AMC 12A Problems/Problem 7"
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==Problem== | ==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? | + | Let <math>a, b, c, d</math>, and <math>e</math> be five consecutive terms in an arithmetic sequence, and suppose that <math>a+b+c+d+e=30</math>. Which of <math>a, b, c, d,</math> or <math>e</math> can be found? |
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| + | <math>\textrm{(A)} \ a\qquad \textrm{(B)}\ b\qquad \textrm{(C)}\ c\qquad \textrm{(D)}\ d\qquad \textrm{(E)}\ e</math> | ||
==Solution== | ==Solution== | ||
| − | + | Let <math>f</math> be the common difference between the terms. | |
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| + | * <math>a=c-2f</math> | ||
| + | * <math>b=c-f</math> | ||
| + | * <math>c=c</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{C}</math>. | ||
==See also== | ==See also== | ||
| − | + | {{AMC12 box|year=2007|num-b=6|num-a=8|ab=A}} | |
| − | + | [[Category:Introductory Algebra Problems]] | |
| − | + | {{MAA Notice}} | |
Latest revision as of 15:40, 5 April 2024
Problem
Let 
, and 
be five consecutive terms in an arithmetic sequence, and suppose that 
. Which of 
or can be found?
Solution
Let 
be the common difference between the terms.
, so 
. But we can't find any more variables, because we don't know what is. So the answer is 
.
See also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
