1999 AHSME Problems/Problem 20

Problem

The sequence $a_{1},a_{2},a_{3},\ldots$ statisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ , $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$ .

Let $m$ be the arithmetic mean of $a_1$ and $a_2$ . We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$ .

By definition, $a_3=m$ .

Next, $a_4$ is the mean of $m-x$ , $m+x$ and $m$ , which is again $m$ .

Realizing this, one can easily prove by induction that $\forall n\geq 3;~ a_n=m$ .