Given $a_1,a_{100}, a_i=a_{i-1}a_{i+1}$, what's $a_1+a_2$? I've been given the following puzzle

Let $a_1, a_{100}$ be given real numbers. Let $a_i=a_{i-1}a_{i+1}$ for $2\leq i \leq 99$. Further suppose that the product of the first $50$ is $27$, and the product of all the $100$ numbers is also $27$.
Find $a_1+a_2$.

I tried the following, looking at the sequence for a moment we see:
$$
a_2=a_1\, a_3\\
a_3=a_2\,a_4\\
\vdots\\
a_{99}=a_{98}\,a_{100}
$$
So $$a_2=\frac 1 {a_{99}} \prod_i a_i =\frac {27}{a_{99}}$$
Looking at the other elements, we find that $$a_3=\frac {a_2} {a_1}, a_4=\frac {a_2}{a_1 a_2}, a_5=\frac {a_2}{a_1a_2a_3},\dots , a_n=\frac {a_2}{\prod_{i=1}^{n-2} a_i}$$
Then, $$27=\prod_i a_i=\prod_{1\leq i\leq 100} \frac {a_2}{\prod_{k=1}^{i-1}a_i}$$
The last product I believe is much more complicated than what I should have gotten... 
Has anyone ran into this puzzle before?

Such a sequence is periodic with period length 6, one cycle
$$ A, B, \frac{B}{A},  \frac{1}{A},  \frac{1}{B},  \frac{A}{B} $$
Note that the product of all six is $1.$  Therefore the product of the first 96 is one, and 27 is $\frac{B^2}{A}.$ The product of the first 48 is one, and 27 is also $AB.$ That is
$$ 27 A = B^2, $$
$$  27  = AB.$$
From $$ 729 = 27AB = (27A)B = B^3$$ I get $729 = B^3 $ and
$$ B = 9. $$ Then
$$ A = 3. $$
The sum is $$ A + B =12.  $$
The cycle of length six becomes
$$  3, \; 9,  \; 3, \;  \frac{1}{3}, \;  \frac{1}{9}, \;  \frac{1}{3}.  $$
Putting two together shows
$$  3, \; 9,  \; 3, \;  \frac{1}{3}, \;  \frac{1}{9}, \;  \frac{1}{3}, \; 3, \; 9,  \; 3, \;  \frac{1}{3}, \;  \frac{1}{9}, \;  \frac{1}{3}  $$
In the original notation, with cycles of six multiplying to one, we have
$$ a_{49} a_{50} = a_1 a_2 = 3 \cdot 9 = 27.  $$
Also
$$ a_{97} a_{98} a_{99} a_{100} = a_1 a_2 a_3 a_4 = 3 \cdot 9 \cdot  3 \cdot  \frac{1}{3}  = 27.  $$
