Geometric Series

Not a mistake: an $r$ that satisfies the above is $$r=\frac{1}{2}\left(\sqrt{61}-1\right)$$ (see http://www.wolframalpha.com/input/?i=4%2B4r%5E2%2B4r%3D64). You can check that this is indeed correct by \begin{align*} a & =4\\ ar & =4\cdot\frac{1}{2}\left(\sqrt{61}-1\right)=2\sqrt{61}-2\\ ar^{2} & =4\cdot\left(\frac{1}{2}\left(\sqrt{61}-1\right)\right)^{2}=62-2\sqrt{61}\\ a+ar+ar^{2} & =4+2\sqrt{61}-2+62-2\sqrt{61}=64 \end{align*}