Problem:
The sequence a1,a2,a3,… satisfies a1=19,a9=99, and, for all n≥3,an is the arithmetic mean of the first n−1 terms. Find a2.
Answer Choices:
A. 29
B. 59
C. 79
D. 99
E. 179
Solution:
For n≥3,
an=n−1a1+a2+⋯+an−1
Thus ( n−1 ) an=a1+a2+⋯+an−1. It follows that
an+1=na1+a2+⋯+an−1+an=n(n−1)⋅an+an=an,
for n≥3. Since a9=99 and a1=19, it follows that
99=a3=219+a2
and hence that a2=179. (The sequ is 19,179,99,99,….)