已知等比数列 $\left\{a_{n}\right\}$ 中, $a_{1}=3$, 且 $4 a_{1}, 2 a_{2}, a_{3}$ 成等差数列, 则 $a_{3}+a_{4}+a_{5}=(\quad)$
由题意可设公比为 $q$, 则 $4 a_{2}=4 a_{1}+a_{3}$,

又 $a_{1}=3, \therefore q=2$.

$\therefore a_{3}+a_{4}+a_{5}=a_{1} q^{2}\left(1+q+q^{2}\right)=3 \times 4 \times(1+2+4)=84$.