Finding sum of terms in a sequence

Question

A sequence has $a_1=-2$ and $a_2=4$ and in the sequence, when $n>2$, $a_{n}$= $\dfrac {a_{n-1}}{a_{n-2}}$. Find the sum of the first $99$ terms of sequence.

I tried to deduce from the patterns produced but couldn't draw any fruitful conclusions. $a_3=-2$, $a_4=-0.5$, $a_5=0.25$ and so on...

Answer 1

Every 6 terms are recurring, thus add first 6 terms and then the sum of first 99 terms is (16x6)+Sum of three terms

Seems -12 is the answer.

Answer 2

HINT

We have that

$$a_{n}= \frac {a_{n-1}}{a_{n-2}}=\frac {a_{n-2}}{a_{n-3}}\frac {1}{a_{n-2}}=\frac {1}{a_{n-3}}=\ldots=a_{n-6}$$

Answer 3

If you compute the first $8$ terms, it will be evident that the sequence repeats in blocks of length $6$.

Summing the first $6$ terms, and then multiplying by $16$ yields the sum of the first $96$ terms.

To finish, add the three remaining terms (which are the same as the first three terms).