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Difference between revisions of "1992 AIME Problems/Problem 8"

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== Problem ==
 
== Problem ==
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For any sequence of real numbers <math>A=(a_1,a_2,a_3,\ldots)</math>, define <math>\Delta A^{}_{}</math> to be the sequence <math>(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)</math>, whose <math>n^\mbox{th}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of thet sequence <math>\Delta(\Delta A^{}_{})</math> are <math>1^{}_{}</math>, and that <math>a_{19}=a_{92}^{}=0</math>. Find <math>a_1^{}</math>.
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For any sequence of real numbers <math>A=(a_1,a_2,a_3,\ldots)</math>, define <math>\Delta A^{}_{}</math> to be the sequence <math>(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)</math>, whose <math>n^\mbox{th}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence <math>\Delta(\Delta A^{}_{})</math> are <math>1^{}_{}</math>, and that <math>a_{19}=a_{92}^{}=0</math>. Find <math>a_1^{}</math>.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Since the second differences of the sequence is constant, the sequence can be expressed explicitly as a quadratic in terms of <math>n</math>.
 +
 
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Since <math>a_{19}=a_{92}^{}=0</math>, and the second differences are all <math>1</math>, <math>a_n=\frac{1}{2!}(n-19)(n-92)</math>.
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Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=7|num-a=9}}
 
{{AIME box|year=1992|num-b=7|num-a=9}}

Revision as of 22:37, 27 February 2008

Problem

For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$, define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$, whose $n^\mbox{th}_{}$ (Error compiling LaTeX. Unknown error_msg) term is $a_{n+1}-a_n^{}$. Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$, and that $a_{19}=a_{92}^{}=0$. Find $a_1^{}$.

Solution

Since the second differences of the sequence is constant, the sequence can be expressed explicitly as a quadratic in terms of $n$.

Since $a_{19}=a_{92}^{}=0$, and the second differences are all $1$, $a_n=\frac{1}{2!}(n-19)(n-92)$.

Thus, $a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}$.

See also

1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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