# 1999 AHSME Problems/Problem 20

## Problem

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

$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$

## Solution 1

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 $\forall n\geq 3;~ a_n=m$.

It follows that $m=a_9=99$. From $19=a_1=m-x$ we get that $x=80$. And thus $a_2 = m+x = \boxed{(E) 179}$.

## Solution 2

Let $a_1=a$ and $a_2=b$. Then, $a_3=\frac{a+b}{2}$, $a_4=\frac{a+b+\frac{a+b}{2}}{3}=\frac{a+b}{2},$ and so on.

It can be observed that for $a_n=\frac{a+b}{2},$ for all $a_n\geq{3}.$

Since $a_9=\frac{a_1+a_2}{2}=\frac{a+b}{2}=99, a+b=198.$ We also know that $a_1=a=19.$

Subtracting $a_1$ from $198,$ we get $b=a_2=\boxed{(E) 179}.$

~Benedict T (countmath1)