Problem:
Suppose that Suppose that {anβ} is an arithmetic sequence with
a1β+a2β+β―+a100β=100
and
a101β+a102β+β―+a200β=200
What is the value of a2ββa1β?
Answer Choices:
A. 0.0001
B. 0.001
C. 0.01
D. 0.1
E. 1
Solution:
Let d=a2ββa1β. Then ak+100β=akβ+100d, and
a101β+a102β+β―+a200ββ=(a1β+100d)+(a2β+100d)+β¦+(a100β+100d)=a1β+a2β+β¦+a100β+10,000dβ
Thus 200=100+10,000d and d=10,000100β=0.01.
Answer: Cβ.
The problems on this page are the property of the MAA's American Mathematics Competitions