Difference between revisions of "1994 AHSME Problems/Problem 6"
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<math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3 </math> | <math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3 </math> | ||
==Solution== | ==Solution== | ||
+ | We work backwards to find <math>a</math>. | ||
+ | |||
+ | <cmath> d+0=1\implies d=1</cmath> | ||
+ | <cmath>c+1=0\implies c=-1</cmath> | ||
+ | <cmath> b+(-1)=1\implies b=2</cmath> | ||
+ | <cmath>a+2=-1\implies a=\boxed{\textbf{(A)}-3.}</cmath> | ||
+ | |||
+ | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME box|year=1994|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:27, 9 January 2021
Problem
In the sequence each term is the sum of the two terms to its left. Find .
Solution
We work backwards to find .
--Solution by TheMaskedMagician
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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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