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

Revision as of 04:11, 18 February 2018

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 $a$ .

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

@@ 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]