jmc

Let the common ratio of the geometric sequence be $r$. We have the equations $140\cdot r=a$ and $a\cdot r=\frac{45}{28}$. In the first equation, we solve for $r$ to get $r=\frac{a}{140}$, and substitute this into the second equation to eliminate $r$, resulting in $a\cdot \frac{a}{140}=\frac{45}{28}$, or $a=\boxed{15}$.