Difference between revisions of "1994 AHSME Problems/Problem 6"

Latest revision as of 16:27, 9 January 2021

In the sequence $..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...$ each term is the sum of the two terms to its left. Find $a$ .

$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$

We work backwards to find $a$ .

$d+0=1\implies d=1$ $c+1=0\implies c=-1$ $b+(-1)=1\implies b=2$ $a+2=-1\implies a=\boxed{\textbf{(A)}-3.}$

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

@@ Line 8: / Line 8: @@
 We work backwards to find <math>a</math>.
-<cmath>d+0=1&\implies d=1\\ c+1=0&\implies c=-1\\ b+(-1)=1&\implies b=2\\ a+2=-1&\implies a=\boxed{\textbf{(A) }-3.}</cmath>
+<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}}