Geometric Series

Question

The first term in a geometric series is 4 and the sum of the first three terms is 64. Find the sum of the first eight terms of the series.

I know the a value is 4, but I'm unsure of how to find the r value.

This is what i have tried:

$a+ ar+ ar^2 = 64$

$4r^2+4r-60=0$

This doesn't give an integer answer, so i doubt that it is the r value. Any hints as to where i went wrong?

Answer 1

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*}

Answer 2

$$r^2+r-15=0\implies r=\frac{-1\pm\sqrt{61}}{2}$$

$$\begin{align}1+r+r^2&=1+\left(\frac{-1\pm\sqrt{61}}{2}\right)+\left(\frac{-1\pm\sqrt{61}}{2}\right)^2\\ &=1+\left(\frac{-1\pm\sqrt{61}}{2}\right)+\frac{31}{2}\mp\frac{\sqrt{61}}{2}\\ &=16\end{align}$$

Thus $a+ar+ar^2=a(1+r+r^2)=4\cdot16=64$