smc

We can setup our first equation as $\frac{3n(3n+1)}{2}=\frac{n(n+1)}{2}+150$ Simplifying we get $9n^2+3n=n^2+n+300\Rightarrow 8n^2+2n-300=0\Rightarrow 4n^2+n-150=0$ So our roots using the quadratic formula are $\frac{-b\pm \sqrt{b^2-4ac}}{2a}\Rightarrow \frac{-1\pm \sqrt{1^2-4\cdot (-150)\cdot 4}}{2\cdot 4}\Rightarrow \frac{-1\pm \sqrt{1+2400}}{8}\Rightarrow 6,-25/4$ Since the question said positive integers, $n=6$, so $4n=24$ $\frac{24\cdot 25}{2}=300$ $\boxed{300}$