# 1992 AIME Problems/Problem 8

## 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. ! Missing { inserted.) 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

Note that the $\Delta$s are reminiscent of differentiation; from the condition $\Delta(\Delta{A}) = 1$, we are led to consider the differential equation $$\frac{d^2 A}{dn^2} = 1$$ This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; $$a_{n} = \frac{1}{2}(n-19)(n-92)$$ as we must have roots at $n = 19$ and $n = 92$.

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

## Solution 2

Let $\Delta^1 A=\Delta A$, and $\Delta^n A=\Delta(\Delta^{(n-1)}A)$.

Note that in every sequence of $a_i$, $a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...$

Then $a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...$

Since $\Delta a_1 =a_2 -a_1$, $a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}$ $a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153$ $a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095$

Solving, $a_1=\boxed{819}$.

## Solution 3

Write out and add first $k-1$ terms of the second finite difference sequence: $a_3+a_1-2*a_2=1$ $a_4+a_2-2*a_3=1$ $a_k + a_{k-2} - 2*a_{k-1} = 1$ $a_{k+1} + a_{k-1} - 2*a_k = 1$

Adding the above $k-1$ equations we get: $\boxed{(a_{k+1} - a_k) = k-1 + (a_2-a_1)} --- (1)$

Now taking sum $k = 1$ to $18$ in equation $(1)$ we get: $18*(a_1-a_2) - a_1 = 153 --- (2)$

Now taking sum $k = 1$ to $91$ in equation $(1)$ we get: $91*(a_1-a_2) - a_1 = 4095 --- (3)$ $18* (3) - 91*(2)$ gives $a_1=\boxed{819}$.

Kris17

## See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS