Problem - 3170
Suppose all the terms in a geometric sequence $\{a_n\}$ are positive. If $|a_2-a_3|=14$ and $|a_1a_2a_3|=343$, find $a_5$.
Firstly, $a_1a_2a_3=a_2^3\implies a_2^3=343 \implies a_2=7$.
Secondly, $|a_2-a_3|=14 \implies |7-a_3|=14 \implies a_3 = 21\quad (\because a_3 > 0)$.
Therefore the common ratio of this sequence equal $a_3/a_2 = 3$. It follows $a_5=a_3\times 3^2 = \boxed{189}$.