Problem:
Two geometric sequences a1,a2,a3,… and b1,b2,b3,… have the same common ratio, with a1=27,b1=99, and a15=b11. Find a9.
Solution:
Let r be the shared common ratio of the two sequences. Because the sequences are geometric, 27r15−1=a15=b11=99r11−1, which implies that r4=311. It follows that a9=27r8=27(311)2=3⋅121=363.
The problems on this page are the property of the MAA's American Mathematics Competitions