[Algebra]
Problem: Suppose that $a$ and $r$ are real numbers. A geometric sequence with first term $a$ and common ratio $r$ has 4 terms. The sum of this geometric sequence is $6+6 \sqrt{2}$. A second geometric sequence has the same first term $a$ and the same common ratio $r$, but has 8 terms. The sum of this second geometric sequence is $30+30 \sqrt{2}$. Determine all possible values for $a$.

(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, $3,-6,12,-24$ are the first four terms of a geometric sequence.)

Solution: Since the sum of a geometric sequence with first term $a$, common ratio $r$ and 4 terms is $6+6 \sqrt{2}$, then

$$
a+a r+a r^{2}+a r^{3}=6+6 \sqrt{2}
$$

Since the sum of a geometric sequence with first term $a$, common ratio $r$ and 8 terms is $30+30 \sqrt{2}$, then

$$
a+a r+a r^{2}+a r^{3}+a r^{4}+a r^{5}+a r^{6}+a r^{7}=30+30 \sqrt{2}
$$

But

$$
\begin{aligned}
a+a r & +a r^{2}+a r^{3}+a r^{4}+a r^{5}+a r^{6}+a r^{7} \\
& =\left(a+a r+a r^{2}+a r^{3}\right)+r^{4}\left(a+a r+a r^{2}+a r^{3}\right) \\
& =\left(1+r^{4}\right)\left(a+a r+a r^{2}+a r^{3}\right)
\end{aligned}
$$



Therefore,

$$
\begin{aligned}
30+30 \sqrt{2} & =\left(1+r^{4}\right)(6+6 \sqrt{2}) \\
\frac{30+30 \sqrt{2}}{6+6 \sqrt{2}} & =1+r^{4} \\
5 & =1+r^{4} \\
r^{4} & =4 \\
r^{2} & =2 \quad\left(\text { since } r^{2}>0\right) \\
r & = \pm \sqrt{2}
\end{aligned}
$$

If $r=\sqrt{2}$,

$a+a r+a r^{2}+a r^{3}=a+\sqrt{2} a+a(\sqrt{2})^{2}+a(\sqrt{2})^{3}=a+\sqrt{2} a+2 a+2 \sqrt{2} a=a(3+3 \sqrt{2})$

Since $a+a r+a r^{2}+a r^{3}=6+6 \sqrt{2}$, then $a(3+3 \sqrt{2})=6+6 \sqrt{2}$ and so $a=\frac{6+6 \sqrt{2}}{3+3 \sqrt{2}}=2$.

If $r=-\sqrt{2}$,

$a+a r+a r^{2}+a r^{3}=a-\sqrt{2} a+a(-\sqrt{2})^{2}+a(-\sqrt{2})^{3}=a-\sqrt{2} a+2 a-2 \sqrt{2} a=a(3-3 \sqrt{2})$

Since $a+a r+a r^{2}+a r^{3}=6+6 \sqrt{2}$, then $a(3-3 \sqrt{2})=6+6 \sqrt{2}$ and so

$$
a=\frac{6+6 \sqrt{2}}{3-3 \sqrt{2}}=\frac{2+2 \sqrt{2}}{1-\sqrt{2}}=\frac{(2+2 \sqrt{2})(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}=\frac{2+2 \sqrt{2}+2 \sqrt{2}+4}{1-2}=-6-4 \sqrt{2}
$$

Therefore, the possible values of $a$ are $a=2$ and $a=-6-4 \sqrt{2}$.

An alternate way of arriving at the equation $1+r^{4}=5$ is to use the formula for the sum of a geometric sequence twice to obtain

$$
\frac{a\left(1-r^{4}\right)}{1-r}=6+6 \sqrt{2} \quad \frac{a\left(1-r^{8}\right)}{1-r}=30+30 \sqrt{2}
$$

assuming that $r \neq 1$. (Can you explain why $r \neq 1$ and $r^{4} \neq 1$ without knowing already that $r= \pm \sqrt{2}$ ?)

Dividing the second equation by the first, we obtain

$$
\frac{a\left(1-r^{8}\right)}{1-r} \cdot \frac{1-r}{a\left(1-r^{4}\right)}=\frac{30+30 \sqrt{2}}{6+6 \sqrt{2}}
$$

which gives

$$
\frac{1-r^{8}}{1-r^{4}}=5
$$

Since $1-r^{8}=\left(1+r^{4}\right)\left(1-r^{4}\right)$, we obtain $1+r^{4}=5$. We then can proceed as above.

Final answer: $a=2$, $a=-6-4 \sqrt{2}$