Problem:
If the sum of the first 10 terms and the sum of the first 100 terms of a given arithmetic progression are 100 and 10, respectively, then the sum of the first 110 terms is
Answer Choices:
A. 90
B. −90
C. 110
D. −110
E. −100
Solution:
The formula for the sum Sn of n terms of an arithmetic progression, whose first term is a and whose common difference is d, is 2Sn=n(2a+(n−1)d). Therefore,
200202S110=10(2a+9d)=100(2a+99d)=110(2a+109d).
Subtracting the first equation from the second and dividing by 90 yields 2a+109d=−2. Hence, 2S110=110(−2), so S110=−110.