Given that the sum of the first 3 terms of a geometric sequence $\{a_n\}$ is $168$, and $a_2 - a_5 = 42$, then $a_6 =$ A. $14$ B. $12$ C. $6$ D. $3$
Given that the sum of the first 3 terms of a geometric sequence $\{a_n\}$ is $168$, and $a_2 - a_5 = 42$, then $a_6 =$\\
A. $14$\\
B. $12$\\
C. $6$\\
D. $3$