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