1994 AHSME Problems/Problem 6: Difference between revisions
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | --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
![\[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\]](http://latex.artofproblemsolving.com/d/d/b/ddb98ab5cb4c79e636eebd6161fcdfea100f7c3c.png)
each term is the sum of the two terms to its left. Find 
.
Solution
We work backwards to find .
![\[d+0=1\implies d=1\]](http://latex.artofproblemsolving.com/9/c/7/9c724e486f574487b2413594589cfb1a73d8a632.png)
![\[c+1=0\implies c=-1\]](http://latex.artofproblemsolving.com/a/4/9/a499bb1932a96cd062ebfa9e1fa6f9b0bd3f502f.png)
![\[b+(-1)=1\implies b=2\]](http://latex.artofproblemsolving.com/3/7/6/37674399b748d7e4b60e93a9790d9169e56121d8.png)
![\[a+2=-1\implies a=\boxed{\textbf{(A)}-3.}\]](http://latex.artofproblemsolving.com/6/e/c/6ecfa8e3b9d5460463735971957381c83183ff10.png)
--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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
