1994 AHSME Problems/Problem 6

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Problem

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 $a$.

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

Solution

We work backwards to find $a$.

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

--Solution by TheMaskedMagician

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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