If $a_n=\frac{a_{n-1}\times a_{n-2}}{2},a_5=-6,a_6=-18$, what is value of $a_3$?

According to the formula, $a_3=\frac{a_2\times a_1}{2}$. But you aren't given $a_1$ or $a_2$. Istead, you are given $a_5$ and $a_6$. You have to work backwards from the fifth and sixth terms of the sequence to find the third term.

Notice what happens if you plug $n=6$ into the formula:

If you plug in the values of $a_5$ and $a_6$,you can solve for the value of $a_4$:

Now you can use the fourth and fifth terms of the sequence to solve for $a_3$: