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 — click to reveal
Let r be the common ratio of both sequences. Then a15=a1r14=27r14 and b11=b1r10=99r10, so we have 27r14=99r10⟹r4=2799=311.Then a9=a1r8=27r8=27(311)2=363.