Problem:
The numbers log(a3b7),log(a5b12), and log(a8b15) are the first three terms of an arithmetic sequence, and the 12th term of the sequence is log(bn). What is n?
Answer Choices:
A. 40
B. 56
C. 76
D. 112
E. 143
Solution:
The first three terms of the sequence can be written as 3loga+7logb,5loga+12logb, and 8loga+15logb. The difference between consecutive terms can be written either as
(5loga+12logb)−(3loga+7logb)=2loga+5logb
or as
(8loga+15logb)−(5loga+12logb)=3loga+3logb
Thus loga=2logb, so the first term of the sequence is 13logb, and the difference between consecutive terms is 9logb. Hence the 12th term is
(13+(12−1)⋅9)logb=112logb=log(b112)
The problems on this page are the property of the MAA's American Mathematics Competitions